Download FE1 MOTION

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FE1
MOTION
OBJECTIVES
Aims
From this chapter you should develop your understanding of the ways that physicists describe
motion. You will appreciate the differences between the everyday meanings of common words like
velocity and acceleration and their scientific meanings - which can be quite different! As well as
learning this new word-language you should also learn how graphs can be used to describe the
quantitative details of motion. You will come to appreciate, we hope, that such graphs are much more
useful and comprehensive than memorised formulas and equations - which have limited validity in
any case.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
position, displacement, vector, scalar, component, time, time interval, average velocity,
velocity [instantaneous velocity], speed, acceleration, centripetal acceleration.
2.
Plot, interpret and find values of kinematic quantities from graphs of displacement component,
velocity component, acceleration component, distance travelled and speed, all plotted against
time interval. Solve kinematic problems using graphical techniques.
3.
State, apply and explain the relation among centripetal acceleration, radius and speed.
4.
State the relation between acceleration and total force and apply the relation qualitatively to
some simple examples.
PRE-LECTURE
Introduction
This chapter deals with ways of describing the motion of objects. The basic concepts are introduced
using examples involving motion in only one dimension (i.e. motion along a straight line). These
ideas are then extended to treat motion in two and three dimensions.
1-1
POSITION AND DISPLACEMENT
A basic description of the motion of an object can be developed using the ideas of position and
time. As time goes on a moving object changes its position relative to some reference point. As the
object moves it also traces out a path, which is usually curved and whose length, measured along the
path, is the distance travelled by the object.
To illustrate the idea of position, consider a car journey from Lane Cove to the University via
the Gladesville Bridge (figure 1.1). The car is continually changing its speed and direction of travel.
(Remember that going up and down hills also involves changing direction.)
FE1: Motion
Lane Cove
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Initial position
Position
Gladesville
Town Hall
University
Figure 1.1
Final position
Position vectors
At any instant, we can specify the position of the car from some reference point, the Sydney
Town Hall for instance. The position could be described by the length and direction of an imaginary
arrow, the car's position vector, drawn from the Town Hall to the car.
However it is usually much more convenient to describe position using a set of rectangular
coordinates, x ,y, z. These three quantities are also called components. In this example x could be
the easterly component of position, y the northerly component, and z the altitude component. In
figure 1.2 x has a negative value (-6.3 km, for example) because the car is to the west, rather than the
east, of the origin (Town Hall).
Lane Cove
Position
( x,y )
2.1 km
Town Hall
6.3 km
University
Figure 1.2
Coordinates or components of position
The components of position are scalar quantities, which can have positive or negative values.
A change in position is called a displacement. It can be specified by quoting the changes in
each of the three position components. Notice that the magnitude of the displacement and distance
travelled are quite different quantities.
Lane Cove
Displacement
Distance
travelled
Town Hall
University
Figure 1.3
Displacement and distance travelled
Displacement, or change in position, is a vector quantity. Distance travelled is a scalar quantity which
cannot be negative.
FE1: Motion
1-2
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VELOCITY IN ONE DIMENSION
In order to develop the other concepts used to describe motion - velocity and acceleration - we can
use one-dimensional examples, with a single position component. The treatment of the other two
components in more general cases is exactly the same.
Example 1.1: One-dimensional motion of a car
Consider a car rolling downhill - out of gear. Let's call its position component x and let the time interval
since it started to roll be t. Suppose that its motion is described by the following table:
time
position component
t /s
x /m
0.0
0
1.0
1
2.0
4
3.0
9
4.0
16
5.0
25
6.0
36
Q1.1
It is often more revealing to display this information pictorially on a graph. Plot the position component
as a function of time elapsed on the graph paper below (figure 1.4).
Position
coordinate
40
x/m
30
20
10
0
0
1
2
3
4
Elapsed time
5
6
t/s
Figure 1.4
A position-time graph.
You should plot the graph yourself using the data tabulated above.
We define average velocity over some time interval:
change in position
average velocity =
.
time taken
In the general case we would need three components to describe the position, so we would
need three components to describe the velocity. In this example we have only one position
component and one component of average velocity:
change in x
average velocity component = change in t ;
∆x
—
or
vx
= ∆t
... (1.1)
(The bar on top of the symbol v indicates 'average' and the subscript x indicates the component of
velocity in the direction corresponding to increasing values of coordinate x.)
The SI unit of velocity is the metre per second, symbol m.s-1.
Q1.2
Use the data of Q1.1 to calculate the component of average velocity over the time interval between 2.0 s
and 5.0 s.
The instantaneous velocity (or just velocity if the context is clear) at any instant is the rate
at which an object's position is changing at that instant. For a one-dimensional motion, the velocity
FE1: Motion
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component is given by the slope of the position-time graph. The slope of a graph at a point is equal
to the slope of the tangent to the graph at that point, which can be found as shown in figure 1.5.
Position
component
Tangent to
graph at
time t
Δx
Δt
t
Figure 1.5
Time
Finding instantaneous velocity from position
Construct the tangent to the position-time curve at the time (t) of interest. Draw a triangle with the
∆x
tangent as hypotenuse. Read or measure off the other two sides and calculate the ratio
.
∆t
Q1.3
Use this definition and the data of Q1.1 to estimate the velocity component of the car at t = 2 .0 s and at
t = 5.0 s. Note that these values are different from the average velocity calculated in Q1.2.
In our one-dimensional example the component of instantaneous velocity is given by the
time-derivative of the position component:
dx
vx = dt
... (1.2) .
In general, velocity is a vector quantity because it has direction. Note that the three
components of a velocity are scalar quantities. Components themselves do not have direction, but
they can take positive or negative values.
In physics the magnitude of a velocity is called speed and, since it is a magnitude, a speed
cannot be negative.
1-3
ACCELERATION IN ONE DIMENSION
By analogy with the definition of average velocity the average acceleration of an object during
some time interval is defined as:
change in velocity
average acceleration =
.
time taken
This definition can be expressed in terms of components as
∆vx
—
ax =
... (1.3)
∆t
The instantaneous acceleration at any instant is defined as the rate at which the velocity is
changing at that instant. Acceleration is also a vector quantity so, in general, we need three
components to describe it. The component of instantaneous acceleration is given by the slope
of the tangent to the velocity-time graph (figure 1.6).
FE1: Motion
Velocity
component,
11
vx
∆vx
∆vx
ax =
∆t
∆t
Figure 1.6
t
Time
Finding acceleration from a velocity-time graph
Measure the slope of the tangent to the curve at the time of interest.
In calculus notation,
dvx
... (1.4)
dt ,
i.e. acceleration is the derivative of velocity with respect to time.
The SI unit of acceleration is the metre per second per second, symbol m.s-2.
To find the acceleration in example 1.1, we would have to find the velocity components at a
series of instants, plot these on a velocity-time graph and find the slopes at various times.
ax
Q1.4
=
Plot the velocity component of the car (example 1.1) at t = 2.0 s and t = 5.0 s on the graph paper below
(figure 1.7).
Velocity
component
v /m.s -1
0
1
Figure 1.7
2
3
4
Elapsed time
5
6
t/s
A velocity-time graph
You should use data from figure 1.4 to construct this graph.
In this example the acceleration was constant so a straight line through the two points you have just
plotted represents the velocity component at all times. Use this graph to find the constant value of the
acceleration component.
FE1: Motion
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LECTURE
1-4
MOTION IN ONE DIMENSION
In the following one-dimensional example the velocity component remains positive, which means
that the direction of travel remains constant. In this special case the distance travelled happens to be
equal to the magnitude of the displacement, or change in position. Remember, however, that this is
not true in general.
Example 1.2: Starting a car
a)
A car is started by rolling it downhill out of gear.
b)
When it is going sufficiently fast the gears are engaged. This starts the engine successfully and the
car is driven to the bottom of the hill
c)
The car is then stopped.
The following position-time graph (figure 1.8) is obtained.
Figure 1.8
Position-time graph for starting a car (example 1.2)
By estimating slopes in figure 1.8 we obtain the velocity-time graph in figure 1.9.
Velocity
component / m.s -1
3
2
1
0
0
2
Figure 1.9
4
6
8
10
12
14
16
Time/s
Velocity-time graph for example 1.2
Then by estimating slopes on the velocity-time graph (figure 1.9) we obtain the acceleration-time graph
(figure 1.10).
FE1: Motion
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2
Acceleration
-2
component/ m.s
1
0
10
2
4
6
12
14
8
16
Time/s
-1
-2
Figure 1.10
Acceleration-time graph for example 1.2
Example 1.3: A velocity-time graph
Suppose that we had been given velocity-time information instead of position-time information. We can
perform a reverse process and find displacement as a function of time.
time /s
velocity component/m.s-1
0
10
1.0
7
2.0
-2
3.0
-17
4.0
-38
5.0
-65
We want to find the displacement component as a function of time, using the starting point as the
reference point. First, we draw the velocity-time graph, figure 1.11.
Figure 1.11
Velocity-time graph for example 1.3
The displacement component after a certain time is given by the area under the velocity-time graph up
to that time. By counting squares, the area under the curve up to t = 1.00 s, say, is 90 small squares
which represents a displacement of
90 × (1.0 m.s-l) × (0.10 s)
=
9.0 m.
i.e. we find that the displacement after the first second is 9.0 m. We can continue this process and plot a
displacement-time graph (figure 1.12). Note that areas under the time axis in figure 1.11 are to be
subtracted since the body is moving in the opposite direction; values of the velocity component are
negative, so each contribution to the displacement is negative.
FE1: Motion
Figure 1.12
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Displacement calculated from figure 1.11
In calculus terminology this operation is an integration:
x=
t
∫0 vxdt
,
i.e. displacement from the starting point is the integral of the velocity with respect to time.
It is also possible to find velocities from an acceleration-time graph in a similar manner if the
starting velocity is known.
1-5
MOTION IN MORE THAN ONE DIMENSION
So far our discussion has been restricted to one-dimensional motion. In three dimensions directions
are obviously important. Displacement, velocity and acceleration are all quantities which need both
their magnitudes and directions to be specified in order to describe them completely. They are
vectors. At any instant, the velocity is in the direction that the object is moving.
Demonstrations:
Ball thrown into the air
(Draw your own diagrams. They should show the path of the object together with arrows to
represent velocity and acceleration at some instant.)
When the ball is thrown vertically upwards, the acceleration is 9.8 m.s-2 downwards
along the same vertical line, and the problem is one-dimensional.
When the ball is thrown upwards at some angle, the problem is two-dimensional since
there is horizontal motion as well as vertical motion. The acceleration is still 9.8 m.s-2
vertically downwards, but now the acceleration is not in the same direction as the velocity. The
velocity of the ball changes in both magnitude and direction; its path curves.
The acceleration is in the vertically-down direction which means that only the vertical
component of velocity changes. The horizontal component of velocity remains constant.
Uniform circular motion
An object moves with constant speed, v, in a circle of radius R. The instantaneous velocity is along a
tangent to the circle and the direction of the acceleration is along the radius towards the centre of the
circle. This is called a centripetal acceleration. Its magnitude is
v2
a = R .
... (1.5)
In this special case the magnitude of the velocity (i.e. the speed) remains constant and only its
direction changes. The magnitude of the acceleration is also constant but its direction keeps
changing, so that it always points towards the centre of the circle. (If the speed of a circular motion
changes there is a component of acceleration tangential to the path but the radial component of the
acceleration is still described by equation 1.5.)
FE1: Motion
1-6
15
FORCES
Having described motion (kinematics), we can move on to examine the causes of changes in motion
or causes of acceleration (dynamics). These are forces.
Firstly, we have the law of inertia: if there are no forces acting, the motion remains
unchanged. If more than one force is acting and these forces cancel or balance each other, this has
the same effect as no force. (This law is also known as Newton's first law of motion.)
Demonstrations
•
Air track with glider - glider at rest, glider moving with constant velocity.
•
Dry ice puck.
•
Motion in a spacecraft cabin.
Secondly, if there is an unbalanced force acting, there is an acceleration in the direction of the
net force.
Since both magnitude and direction are needed for the complete specification of a force, force
is a vector quantity.
Demonstrations
What is the direction of the force in each of the following cases?
•
A ball thrown into the air.
•
Circular motion examples.
POST-LECTURE
1-7
QUESTIONS
Motion in one dimension
Q1.5 a) Consider the velocity and acceleration-time graph in example 1.2 above. Give an account of how the
velocity and acceleration change during each segment.
b) Consider the velocity-time graph in example 1.3 above. Note that areas between the graph and the time
axis represent positive contributions to the displacement if the curve is above the axis and negative
contributions if the curve is below.
What happens to the displacement when the velocity goes through zero to become negative ?
What can you say about the areas above and below the time axis when the displacement is zero (i.e.
when the object has returned to its starting point)?
What does a negative displacement mean?
Q1.6
Here is a velocity-time graph (figure 1.13) for a car journey along a straight road.
80
Component of
velocity / km.h-1
40
0
-40
-80
0
Figure 1.13
1
2
3
4
5
Time/min
Example: one dimensional motion of a car
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FE1: Motion
Give a qualitative description of the motion of the car during the journey.
How far is the car from its starting point at the end of its journey? [Hint: Accelerations for each
segment are given by slopes, distances are given by areas.]
Q1.7
The driver of a car wants to average 60 km.h-1 for a 20 km trip. The car travels the first 10 km at a
constant speed of 40 km.h-1. How fast must it go over the last 10 km?
Q1.8
This exercise is inspired by the N.S.W. Motor Traffic Handbook. It concerns stopping distances.
Assume that the reaction time to apply the brakes of a car is l.0 s, and assume that the brakes give a
constant acceleration of magnitude 5 m.s-2.
Compare the distances required to stop the car travelling at
(a) 9 m.s-1 and (b) 18 m.s-1. (9 m.s-1 is approximately 32 km.h-1.)
(Hint: use velocity-time graphs.)
Motion in more than one dimension
Q1.9
An object is accelerating if
(a) the magnitude and direction of its velocity both change, or
(b) only the magnitude of its velocity changes, or
(c) only the direction of its velocity changes.
Give examples of each.
Forces
Q1.10
Suppose you are standing in front of a frictionless table and an object on it is moving from left to right at
constant velocity. What happens to the motion if:
(a)
you apply a force directed to the right;
(b)
you apply a force directed to the left;
(c)
you apply a force directed towards yourself?
Q1.11
Consider the motion of an object on which a single force acts. What can you say about the magnitude and
direction of the forces that produce the following effects:
(a)
object moving from left to right with constant velocity;
(b)
object moving from left to right with velocity increasing;
(c)
object moving from left to right irregularly (i.e. stopping and starting);
(d)
object moving in circle at constant speed;
(e)
object moving in circle at variable speed?
APPENDIX
1-8
EXAMPLES OF MATHEMATICAL DESCRIPTIONS OF MOTION
In the lecture you have seen how motion can be described by a table of numbers or by a graph. In
some special situations a mathematical equation can be used.
For example, if an expression for the displacement is given, it may be differentiated to give
velocities and accelerations. Or, if the velocity is given, it may be differentiated to give accelerations.
Conversely, if the acceleration is given, together with some statement about what the moving
object is doing at time t = 0, it may be integrated to give velocities and displacements. Or if the
velocity is given together with some similar statement, it may be integrated to give the displacement.
Q1.12
The displacement of an object is given by
a)
x
=
A + Bt + Ct2 + Dt3 or
b)
x = A cos(ω t ),
where A, B, C, D, ω are constants.
In each case,
i)
find the displacement of the object at time t = 0, and
ii)
differentiate to find the velocity and the acceleration as functions of time.
Q1.13
At t = 0, an object is at position x = x0 and has velocity component v = v0 . Its acceleration is
a)
a
=
c
b)
a = - kt
where c and k are constants.
In each case, integrate to find velocities and displacements at any subsequent time t.
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17
FORCE
FE2
OBJECTIVES
Aims
This chapter will almost certainly require a re-organisation or replacement of your intuitive views
about the physical world! You should aim to understand the physical concept of force which is
probably quite different from the normal intuitive concept that you have already acquired. You will
also learn the principles, known as Newton's laws of motion, which underpin the scientific idea of
force. You will learn to apply the concept of force and Newton's laws to simple examples.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
mass, vector, component, force, newton, fundamental force, total force [resultant force, net
force], weight [gravitational force], gravitational acceleration [gravitational field], equation
of motion [Newton's second law], Newton's third law, centripetal force, contact force,
tension.
2.
Describe the general system of classification of fundamental forces.
3.
State, explain and apply Newton's third law.
4.
Relate tensions in strings or ropes to the forces exerted by those strings or ropes.
5.
Draw diagrams showing the forces acting on objects.
6.
State the equation of motion and apply it to simple problems by considering force components
in mutually perpendicular directions.
7.
Use force-time graphs to find accelerations, velocities and speeds of objects.
PRE-LECTURE
Introduction
Accelerations are caused by unbalanced forces. This idea was discussed in the last chapter. A
mathematical expression of this result, the equation of motion, is the main topic in this chapter.
Revise questions 1.9, 1.10 and 1.11 in FE1.
2-1
COMPONENTS
The idea of components, which was applied to the vector quantities position, velocity and acceleration
in chapter FE1, can be extended to other kinds of vector quantities such as force.
The value of a vector quantity can be described by specifying its magnitude (number × unit)
and its direction. An alternative, equivalent, description is in terms of three components referred to
three mutually perpendicular directions or coordinate axes. The relation between a vector and its
components is most easily illustrated on two-dimensional paper using two dimensions, with two
reference directions.
FE2: Force
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Direction V
a
a sin θ
θ
a cos θ
Figure 2.1
Direction H
Components of a vector
The component of vector a in a direction at an angle θ to the vector's direction is defined to be
equal to a cosθ. In the case illustrated above the component in the direction labelled H is aH = a
cosθ and the component in the direction labelled V is aV = a cos(90° - θ ) or a sinθ . Note that each
component is a scalar quantity (i.e. one which has no direction). The value of a component might
be positive, zero or negative, depending on the value of the angle θ .
2-2
MASS
Mass is a property of an object which is one way of specifying the quantity of matter in it. The
concept of mass can be explained formally in terms of the effect of a total force on a body, using the
equation of motion discussed later in this chapter. Alternatively mass can be determined in terms of
the gravitational force on the object: at a fixed location the gravitational force on different objects
(their weights) are proportional to their masses. The SI unit of mass is the kilogram (symbol kg)
and the unit of weight is the newton (N).
LECTURE
2-3
THE NATURE OF FORCE
Pulls, pushes, compressions, friction, gravity - these are forces that we all are familiar with. All the
forces occurring in nature can be classified into four fundamental types. In decreasing order of
strength these types are:
strong nuclear,
electromagnetic,
weak nuclear,
gravitational.
Nuclear forces
The strong nuclear force holds neutrons and protons together in the nucleus of an atom. It is the
force associated with radioactive alpha decay and nuclear energy, including the atomic bomb and the
hydrogen bomb.
The weak nuclear force holds the neutron together. It is associated with radioactive beta
decay. (Beta decay of carbon-14 can be used for radioactive tracing in biological studies, and for
carbon dating of archaeological items.)
Both types of nuclear force act only over a very short range of distances comparable with the
size of an atomic nucleus, about 10-14 m (10 fm).
Gravity
The gravitational force is the force which causes objects to fall. It is an attractive force between
the Earth and the falling object. This force on a falling object is a special instance of a universal
force which acts between every pair of objects. The fundamental law of gravitational attraction says
FE2: Force
19
that the magnitude of the force between between two particles, with masses m1 and m2 , at a distance
d apart is
m1 m2
F ∝
.
d2
Note that this force extends to infinite distances and is always attractive. The attractive force
between Earth and Moon, Sun and planets, etc is gravitational.
The weight of an object, on the Earth or on the Moon etc, is the gravitational force acting on it.
At any particular place near the Earth's surface, the weight W of a body is proportional to its mass m:
W = mg
... (2.1).
The constant of proportionality g is the magnitude of the Earth's gravitational field but it is often
loosely called the acceleration due to gravity.
Electromagnetic force
The interactions between atoms and the charged particles within atoms are all electromagnetic.
These forces determine whether a substance will be a solid, liquid, gas or a plasma. They are the
forces involved in chemical reactions and biological processes.
The macroscopic forces described as pulls, pushes, friction etc can be explained, on a
molecular scale, as electromagnetic forces. In general, the force between molecules is strongly
repulsive at short distances but weakly attractive at greater distances.
Force
component
REPULSIVE
Equilibrium
Separation
0
ATTRACTIVE
Figure 2.2 Intermolecular force
The equilibrium separation, where the force is zero, is typically about 10
-10 m (0.1 nm).
On a macroscopic scale the forces on electric charges are electromagnetic, as are those
involved in power generation and electronic communications, computers, and measuring instruments.
Electromagnetic forces extend to infinite distances and can be attractive or repulsive.
2-4
PAIRS OF FORCES
Note that all forces, whether they be fundamental or macroscopic, always occur in pairs. For
example, the gravitational force exerted on the Sun by the Earth is equal in magnitude but opposite in
direction to the gravitational force exerted on the Earth by the Sun.
The electrostatic force exerted on a charged particle, A, by another charged particle charge, B,
is equal in magnitude but opposite in direction to the electrostatic force exerted on B by A. The
same is true for the molecular forces between two molecules
This has macroscopic consequences. Consider a block sitting on a table. The block exerts a
downward force on the table and the table exerts an upward force of equal magnitude on the block.
FE2: Force
These are all examples of Newton's third law of motion.
2-5
THE EQUATION OF MOTION
Forces are responsible, not for motion itself, but for changes in the motion of a body. Change in
motion implies acceleration. The effect of the forces is described by the equation of motion, which
is also known as Newton's second law of motion. The equation relates total force F to acceleration
a. For a body whose mass m does not change:
total force = mass × acceleration.
In symbols this equation of motion is
F = ma
... (2.2).
A word about notation: symbols for vector quantities are printed in bold-face type (e.g. F) while
ordinary italic type is used to indicate the magnitude of a vector (e.g. F means |F|.)
Note the following important points about the equation of motion.
•
Only those forces which act on a body affect the motion of that body.
•
All the forces acting must be included in the total force. (Total force is also called net force or
resultant force.)
•
Directions of forces must be considered in the calculation of the total force. (The procedure
for doing this is given below.)
•
The acceleration has the same direction as the total force.
•
The SI unit of force is the newton, symbol N.
Example with one force acting: ball thrown in the air
A ball is thrown into the air (either vertically or at some angle). If the force exerted on the ball by the air
is negligible, then the total force on the ball is equal to the ball's weight, W = mg downwards. Hence the
acceleration a is equal to g, downwards.
Near the earth's surface, g = 9.8 m.s-2.
Uniform circular motion
Consider an object moving in a circle with constant speed v. The total force in this case is called
centripetal force. It follows from the equation of motion (F = ma ) and the formula for
centripetal acceleration (v2 /R), which is towards the centre of the circle, that
mv2
F = R
and the direction of F is towards the centre of the circle.
Note that centripetal force is not a new kind of force; it is just a fancy name for the radial
component of the total force. In the case of uniform circular motion the radial component is the only
component of the total force.
Calculation of total force
When several forces are acting on an object, they must be combined to give the total force. In
general, the forces will be acting in different directions, so combining them presents a problem.
Fortunately the components of the motion in each of any three mutually perpendicular directions
can be treated separately. This separation of the motion into components is helpful even when there
is only one force acting.
For example, if a ball is thrown into the air at some angle to the horizontal, the vertical and
horizontal components of the motion can be treated separately. Provided that air resistance is
negligible, the vertical motion has an acceleration vertically down, but the horizontal motion has
constant horizontal velocity.
Example: a sliding block
A block is being pulled along a frictionless table. The pulling force is directed upwards at an angle θ to
the horizontal.
20
FE2: Force
21
N
P
θ
W
Figure 2.3
Block pulled across a frictionless table
Forces acting on the block:
•
the weight of the block, W, vertically downwards;
•
a contact force, N, exerted by the table on the block, vertically upwards;
•
the pulling force, P.
In this case all the forces act in one plane, the plane of the diagram, so the problem is two-dimensional.
We need only two component directions: vertically up and horizontally to the right are convenient.
Components of the pulling force P:
• horizontal component
PH =
• vertical component
PV
=
P cosθ
P cos (90°-θ )
to the right;
upwards.
Note that the two component forces PH and PV are together equivalent to the pulling force P.
P
PV = Pcos(90° - θ)
is equivalent to
θ
Figure 2.4
PH = Pcos(θ)
Representing a force by its components
The total force
The horizontal component of the total force is FH = PH to the right. So the horizontal component of the
acceleration is equal to
PH
to the right. (Note that components of accelerations are defined in the same
m
way as components of forces.)
The vertical component of the total force FV = N + P V - W, upwards. This must equal zero since the
block does not accelerate vertically. We see therefore that the force the table exerts on the block,
N = W - P V. Note that components, not magnitudes, are added; the component of the weight in the
vertically up direction is -W.
General procedure
Summarizing, the general method is as follows.
(i) Draw in all the forces - make sure that none is omitted.
(ii) Choose three mutually perpendicular directions and calculate the force components along these
directions. (Note that if the motion is confined to a plane as in the example above you need
only two mutually perpendicular directions.)
(iii) Calculate the component of the total force in each direction, by adding the individual
components, and taking account of their signs.
(iv) Calculate the acceleration component in each direction, using the equation of motion.
Note that sometimes you are given the acceleration, in which case you can use the equation of
motion to give the components of the total force.
FE2: Force
22
POST-LECTURE
2-6
QUESTIONS
Fundamental forces
Q2.1 a ) We have said that the gravitational force is weaker than the electromagnetic force. Is this your experience?
Why might the electromagnetic force appear to be weaker than the gravitational force?
b) Nuclear forces were the last to be discovered. Why was this so if they are the strongest?
c) How would you rate the forces in terms of their significance to biological systems?
Molecular forces
Q2.2 a) Why does a block of metal resist changes in its shape when you attempt to compress it, or when you
attempt to stretch it?
b) Suppose an object is hanging from the end of a fine wire. What holds the wire together? If the load is
increased the wire stretches. How do the forces in the wire change?
c) Compare the force required to pull a brick apart with the force required to lift it up. Which is greater?
What does this suggest about the relative strengths of molecular (electromagnetic) force and gravitational
force?
Weight
Q2.3
A spring balance measures the weight of an object, i.e. the gravitational force on it. The gravitational
force on an object near the surface of the Moon is about one-sixth of that near the surface of the Earth.
a) Suppose a particular object has a mass of 2 kg. According to the spring balance, what would its weight be
on Earth? On the Moon? (An approximate answer will do.)
b) How could you use the spring balance to measure the mass of an object?
Equation of motion
Q2.4 a ) An object on the end of a string is being swung in a horizontal circle at constant speed. What provides the
centripetal force?
If you want the object to swing around at twice the speed, how much greater would the centripetal force
have to be?
Sketch the path the object takes if the string breaks.
b) What provides the centripetal force on the various parts of a revolving wheel?
c) What provides the centripetal force on a satellite orbiting about the earth?
The following dynamics problems, where several forces are involved, can all be attacked using
the method outlined in the lecture.
Q2.5
A block, mass 2.00 kg, is being pulled along a frictionless table. The pulling force is 20.00 N in a
horizontal direction (figure 2.5).
F
= 20.00 N
Figure 2.5
Draw in the weight of the block and the force that the table exerts on the block.
Take components of forces in horizontal and vertical directions.
a)
What is the horizontal acceleration?
b)
What is the total vertical force? (Hint: what is the total vertical acceleration?)
c)
Suppose now that there is a horizontal frictional force of 10.00 N opposing the motion. What is the
horizontal acceleration?
FE2: Force
Q2.6 a )
A car is being driven along a flat road.
What are the forces acting on the car?
What difference does it make if the car is moving at constant velocity or accelerating?
b)
The car is rounding a curve.
What provides the centripetal force?
c)
The car is on a hill with slope θ.
What is the downhill component of its weight?
Q2.7
23
A block, mass 2.00 kg, is being pulled along a frictionless table. The pulling force is 20.00 N at an angle
of 30° to the horizontal (figure 2.6).
F = 20.00 N
30°
Figure 2.6
Draw in all the other forces.
a)
b)
What is the total horizontal force? What is the total vertical force?
What is the horizontal acceleration? What is the vertical acceleration? (Will the block be lifted off the
table?)
c)
What would happen if the magnitude of the force were doubled?
d)
Suppose the table were not frictionless, how would this affect your answers?
Q2.8
Two forces are acting on an object (figure 2.7).
4.00 N
y
x
90°
30°
8.00 N
Figure 2.7
If these were the only forces acting, the object would accelerate.
a)
Find the components of the total force in the y ↑ and x → directions.
b)
What additional force or forces would stop the object from accelerating?
FE2: Force
2-7
24
TENSION
When you pull on one end of a rope which is attached to some object at the other end, the rope will
exert a force on that object. Something also happens to the rope itself. To understand this, think
about some place inside the rope and imagine that the rope consists of two parts. Each part exerts a
force on the other part and, according to Newton's third law, those forces have the same size (figure
2.8). This common magnitude of the force between the two parts of the rope is called the tension in
the rope at the particular place considered.
magnitude
Figure 2.8
T
magnitude
T
Internal forces and tension
The force exerted on the object is equal to the tension in the rope at the point where the rope is
joined to it and the force exerted on the rope at the other end is equal to the tension there. In general,
the tension in the rope will be different at different parts of the rope. In some cases where the
applied forces are much larger than the rope's weight, or if the rope is not accelerating, we can make
the useful approximation that the tension is roughly the same throughout the rope. That is the case
in the next problem.
Q2.9
Suppose that your car is stuck in mud at the side of the road. You have the idea that if you tie a rope, one
end to the car and the other end to a tree, and apply a force perpendicular to the rope at the middle, you will
be able to move your car (figure 2.9).
Figure 2.9
a)
How to pull a car out of a bog
Why is this a good idea ?
Hints
• Suppose you have started applying the force to the rope. The rope has taken on the shape shown in
figure 2.10 but the car has not yet started to move. θ is a very small angle.
T
θ
T
P
θ
Figure 2.10
Forces on a point in the rope
• Consider the forces acting on the bend in the rope. The forces at this stage still balance, so take
components to see how the tension compares with the applied force, P. What is the magnitude of the force
on the car? On the tree?
b)
Where might this good idea go wrong ?
FE2: Force
2-8
25
AN ACCELERATING SYSTEM
Consider a person in a lift which has a component of acceleration, vertically upwards, of magnitude a
(figure 2.11). The forces on the person are her weight W = mg downwards and a contact force
exerted by the floor of the lift, upwards.
a
mg
N
Figure 2.11
Forces on a person in an accelerating lift
The person's upward acceleration is equal to the lift's acceleration, a, so
N - mg = ma .
In common usage the term weight has a different meaning from the definition given in the
lecture. Commonly, the weight of an object is the force that it exerts on its support, the floor, etc.
In a lift accelerating upwards, the downward force exerted on the floor by the passenger is
equal to N. Since this is greater than its usual value, mg, the passenger's apparent weight is greater
than normal.
Q2.10 a ) Suppose the lift is accelerating downwards. How does N compare with mg now?
b)
When do you experience "weightlessness"?
c)
Why do you get a strange feeling in your stomach?
Interlude: Times
INTERLUDE 2 - THE RANGE OF TIMES IN THE UNIVERSE
age of the universe
time since earliest humans
1 year
1 day
1 minute
period of a sound wave
time for light to travel 1 m
period of an atomic vibration
period of a nuclear vibration
time/seconds
101 8 __
_
age of the Earth
_
15
10 __
_
_
12
10 __
_
_
109 __ human life span
_
_
6
10 __ 1 month
_
_
3
10 __ 1 hour
_
_
1
__ heart beat
_
_
10-3 __
_
_
-6
10 __ period of a radio wave
_
_
-9
10 __
_
_
-12
10 __
period of a molecular vibration
_
_
10-15__
_
_
18
10 __ time for light to cross an atom
_
_
-21
10 __
_
_
time for light to cross a nucleus
-24
10 __
26
27
FE3
EQUILIBRIUM
OBJECTIVES
Aims
In this chapter you will learn the concepts and principles needed to understand mechanical
equilibrium. You should be able to demonstrate your understanding by analysing simple examples
of equilibrium. You will also learn about the interactions between fluids and solid bodies as well as
the concepts - buoyant force and pressure - used to describe the interactions.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
translational motion, rotational motion, rigid body, equilibrium, stable equilibrium, unstable
equilibrium, neutral equilibrium, axis, torque [moment of a force], centre of gravity,
buoyancy, buoyant force, Archimedes' principle, pressure, pascal, density, barometer.
2.
State and apply the relation between force and torque.
3.
State the conditions for equilibrium and apply them to simple problems.
4.
Describe and explain how the centre of gravity of a body can be located.
5.
Describe and explain the forces acting on a body which is wholly or partly immersed in fluid.
Solve simple problems involving buoyancy.
PRE-LECTURE
Introduction
This chapter deals mainly with the equilibrium of rigid bodies. The conclusions about rigid bodies
can also be applied to some examples of non-rigid bodies, such as bodies of fluid at rest. We start
with two simple examples of objects in equilibrium: an object at rest and one moving with constant
velocity.
All the examples and principles discussed in this chapter are restricted to systems in which all
the forces can be described in two dimensions - a plane. The extension to general three-dimensional
systems uses the same concepts but is mathematically much more complex.
3-1
TRANSLATION AND ROTATION
The simple descriptions of motion that we have used so far implicitly treated the motion of only one
point in a body. That is a good kind of description provided that all points in the body follow
similar, parallel, paths. That kind of motion is called pure translational motion. As well as
translational motion, a body can also execute rotational motion (like that of a spinning wheel) and
vibration (like that of a shaking jelly). A body which cannot vibrate noticeably is said to be rigid; its
shape and size do not change significantly when it is acted on by a system of forces. The most
general kind of motion of a rigid body is, therefore, a combination of translation and rotation; the
flight of a boomerang is a good example.
FE3: Equilibrium
3-2
EQUILIBRIUM OF FORCES
Q3.1
A block is at rest on a table.
Figure 3.1
An object at rest.
28
What are the forces?
a) On figure 3.1 draw in all the forces acting on the block.
The block is not accelerating, so the net force on it must be zero. What does this tell you about the
vertical forces and any horizontal forces that may be present?
b) Suppose that a horizontal force is applied in an attempt to push the block along the table, but the block
does not move. What is opposing the horizontal force?
Q3.2.
Block moving at constant velocity
A block is being pulled across the table at constant velocity (figure 3.2).
pull
Figure 3.2
An object moving with constant velocity
Draw in the other forces.
A frictional force opposes the motion. Draw in this force and the other forces on the block.
Again, the block is not accelerating so the net force on it must be zero. What does this tell you about
the vertical and horizontal forces?
The meaning of equilibrium
The examples in questions 3.1 and 3.2 are simple illustrations of equilibrium. In both cases the
velocity of the block is constant, i.e. its acceleration is zero, and the total force on it is zero.
However it is not enough that the forces balance in order to have equilibrium. This guarantees
only that there is no change of translational motion, i.e. that the motion of the body as a whole does
not change. The definition of equilibrium needs to be extended to include the requirement that the
rotational motion of the body also remains constant. Thus, for example, a body which is completely
at rest is in equilibrium only if it does not start to move or rotate. As another example, a wheel
rotating about a fixed axle, is defined to be in equilibrium only if its rotational speed does not
change.
In order to consider this rotational aspect of equilibrium we need the concept of torque. We
shall then see that for equilibrium, torques, as well as forces, must balance.
FE3: Equilibrium
29
LECTURE
3-3
TORQUE
Example 3.1. Wheel on a fixed axle
Consider a wheel which can rotate about its axle. The axle remains in a fixed position. An
object is hung by a string from the rim of the wheel as shown in figure 3.3.
FA
W
Figure 3.3
F
An unbalanced wheel
When the object is released, the wheel starts to rotate so, clearly, it is not in equilibrium.
The forces acting on the wheel are shown. W is the wheel's weight, FA is the force exerted by
the axle on the wheel and F is the force exerted by the string attached to the falling object,
which causes the wheel to rotate. Note that these forces are not all acting through the same
point.
In this example, rotation occurs about an axis - a line in space - which is perpendicular to the
plane of the forces involved. In such cases we define torque as follows.
The torque of a force F about a specified axis is defined as
τ = Fx
... (3.1)
where x is the perpendicular distance from the axis to the force's line of action. Torque is also
known as the moment of a force. Note that we can, in principle, define many torques for each
force, one for every possible choice of axis. It is not necessary for the axis to be a possible axis of
rotation.
Example 3.1 - continued
x
F
Figure 3.4
Torque on a pivoted wheel
In this example, we consider torques about the wheel's axis of rotation, its axle. Only
the force F has a non-zero torque about the axle so there is a net torque, due to F, acting on the
wheel. This net torque produces an angular acceleration, i.e. it changes the angular speed of
rotation about the axle.
FE3: Equilibrium
30
Demonstrations
•
Consider two wheels which have the same shape and the same total mass. One has a dense
metal rim, the other has a dense metal axle. Identical loads are hung from their rims. This
demonstration shows that the angular acceleration depends on the distribution of mass in the object.
metal rim
metal
axle
Figure 3.5
•
3-4
Effect of mass distribution on angular acceleration
Identical loads were attached to the wheels and then released. The angular acceleration
was greater for the wheel whose mass is more concentrated near the axle.
Other demonstrations show rotational motion when the axis is not fixed.
EQUILIBRIUM OF TORQUES
Example 3.1 - continued
The wheel has no translational motion so the total force on it must be zero.
Forces
Torques
FA
Torque
Fx
W
F
Figure 3.6
Forces and torque on the pivoted wheel
The vertically upward component of the total force = FA - W - F = 0 , but the net torque, Fx, is
unbalanced. Note that this torque produces a clockwise rotation.
Example 3.2: Equilibrium of the wheel
To bring our wheel back into equilibrium, a torque of the same magnitude but in the opposite
(anti-clockwise) sense would have to be provided. Another object with the same mass could be
attached to the opposite side of the wheel in order to achieve this. See figure 3.7. (Note that the
supporting force FA will take on a new value.)
Forces
Torques
FA
-Fx
Fx
W
F
F
Figure 3.7
Forces and torques at equilibrium
FE3: Equilibrium
3-5
31
CONDITIONS FOR EQUILIBRIUM
In general, then, the conditions for equilibrium of an object which is free to rotate about a fixed axis
are:
(i) total force acting on the object = 0;
(ii) total torque about the axis = 0 .
Note that, since force is a vector quantity, the calculation of the net force must take account of
directions. This can be done using the method of components introduced in chapter FE2. Torque,
as defined here, is a scalar quantity whose values need to be associated with either clockwise or
anti-clockwise rotation. We can assign positive values to one of these two senses, and negative
values to the other. (The usual convention makes anti-clockwise values positive.)
Example 3.2 - continued
When we apply these conditions to the wheel with two objects hanging from its rim, we can
choose vertically down as a component direction for the forces (all horizontal force components
are equal to zero).
(i)
F + W + F - FA = 0 .
(FA takes on a new value when the second object is attached.)
The condition for balancing the torques is satisfied because the two torques have the same
magnitudes but opposite senses:
(ii)
Fx - Fx = 0.
It is obviously inconvenient to have to draw two diagrams for each example - one showing
forces and one showing torques. Henceforth both forces and torques will be shown on the same
diagram. Remember, however, that forces and torques are quite different entities and must be
combined separately.
3-6
CENTRE OF GRAVITY
Example 3.3. Centre of gravity of a flat object
A flat object is pivoted at the point P. See figure 3.8. Imagine that the object is divided up into
little pieces. The weight of each piece provides a torque about the pivot. The object will be in
equilibrium only if the torques due to all these pieces add up to zero.
P
Figure 3.8
Distribution of gravitational forces on an object
It is obviously inconvenient to have to consider the weights of all the little pieces of the
object separately. Fortunately, their total can be represented by the total weight, W , acting
through a point called the centre of gravity.
FE3: Equilibrium
32
Torque
Equilibrium
P
Centre
of gravity
P
W
Figure 3.9
W
Centre of gravity and the total weight force
Then W provides a torque about the pivot (equal to the sum of the many small torques).
The object will be in equilibrium if its centre of gravity lies vertically below the fixed
pivot point. In this position the length of the perpendicular from the pivot to the line of action
of the weight is zero, and so is that from the pivot to the supporting force at the pivot.
Therefore the torques of the weight and the supporting force about the pivot are both zero.
Example 3.4. Locating the centre of gravity
To locate the centre of gravity of a flat object, first mark a vertical line showing the line of
action of the weight. The centre of gravity must be somewhere on this line. Then choose a
different pivot point and repeat the process. The centre of gravity must be also be somewhere
on the new line; so it must be at the intersection of the lines.
A wheel is symmetric about its axis and the centre of gravity is at the centre of the
wheel. This is easily verified.
3-7
EQUILIBRIUM OF A SYSTEM OF OBJECTS
Example 3.5: Two children balancing on a seesaw.
This example is analysed in terms of forces and torques acting on a system which consists of the
two children and the plank. The forces acting on this system are the weights of the two
children, the weight of the seesaw's plank and a vertical supporting force at the pivot.
The weights of the children, W 1 and W 2 act at distances x1 and x2 from the pivot, as
shown on the figure. These forces give torques W 1 x 1 (anticlockwise) and W 2 x 2 (clockwise)
respectively about the pivot.
Suppose that the centre of gravity of the plank is directly above the pivot so that the
weight WS of the seesaw plank acts downwards at the pivot. N is the upward supporting force
exerted by the pivot on the plank. Each of these two forces gives zero torque about the pivot.
N
x2
x1
W1 x 1
W 2x 2
W1
Figure 3.10
WS
W2
Balancing on a seesaw
For equilibrium the following conditions must be satisfied.
(i) Total force acting on the system = 0.
Taking force components in the vertically downward direction:
FE3: Equilibrium
33
W 1 + W 2 + W S - N = 0.
(ii)Total torque about an axis through the pivot = 0.
Taking clockwise as the positive sense:
W 2 x 2 - W 1 x 1 = 0.
This analysis is essentially the same as that for a beam balance.
3-8
EQUILIBRIUM OF A FREE OBJECT
A free object is one that is not pivoted. This is a more general situation than the case of a fixed axis.
Demonstration
moves forward
spins anticlockwise
push
push
moves forward
spins clockwise
moves forward
push
no spin
Figure 3.11
Motion of a free object
The centre of gravity is shown as a heavy dot.
A net force not acting through the centre of gravity of a rigid body will cause translational
acceleration of the object as well as change in its rotational motion. The resulting motion can
be described as a combination of translational motion of the centre of gravity and rotational
motion about the centre of gravity.
3-9
GENERAL CONDITIONS FOR EQUILIBRIUM
The general conditions for equilibrium are as follows
(i) The total force must be zero (as before).
(ii) The total torque about any axis must be zero.
In many cases it is convenient to consider torques about axes through the centre of gravity.
3-10
BUOYANCY
When a solid object is wholly or partly immersed in a fluid, the fluid molecules are continually
striking the submerged surface of the object. The forces due to these impacts (which are sometimes
called pressure forces) can be combined into a single force, the buoyant force.
FE3: Equilibrium
Figure 3.12
Forces exerted by a fluid
Note that, for clarity, we show only the forces exerted by the surrounding fluid in this and the
following diagram.
We want to find the magnitude of the buoyant force and the point through which it acts.
Buoyant force on a completely submerged object
Figure 3.13
Buoyant force on a submerged object
To work out how big this buoyant force is, and where it acts, we use the trick of thinking about
a 'block' of fluid, which has exactly the same shape and size as the solid object. This imaginary
portion of fluid is often called the 'displaced fluid'.
Figure 3.14
Fluid "displaced" by the submerged object
The pressure forces on this imaginary displaced fluid are exactly the same as the pressure
forces on the solid object. So their total effect, the buoyant force, will be the same, irrespective of
what is inside the broken outline. The buoyant forces on the solid object and the 'displaced fluid' are
identical.
Now the displaced fluid must be in equilibrium. Since the only other force on it is its weight,
which acts through its centre of gravity, the buoyant force must be equal to its weight, and it must act
vertically upward through its centre of gravity. Hence the buoyant force on the submerged block
must be equal to the weight of the displaced fluid and it must act vertically up through the centre of
gravity of the 'displaced' fluid body.
34
FE3: Equilibrium
35
Buoyant force
Weight of displaced fluid
Figure 3.15
Forces on the displaced fluid
This conclusion is known as Archimedes' principle.
Buoyant force on a partly submerged object
Air
Buoyant force
Liquid
Figure 3.16
Buoyant force on a partly submerged object
Note that there is also a gravitational force (weight).
In this case the object is immersed in two fluids one of which is the air. The diagram shows
only the total force exerted by these fluids on the object.
Consider an imaginary block composed of the two fluids 'displaced' by the object.
Displaced air
Displaced liquid
Figure 3.17
Fluids displaced by the partly submerged object
This fluid block is in equilibrium and, just as before,
buoyant force = total weight of the displaced fluids,
and acts vertically upward through the centre of gravity of the displaced fluids.
Note that, for an object floating in a liquid, the buoyant force due to the displaced air is usually
negligibly small.
FE3: Equilibrium
Example 3.6: A heavy object supported in water by a string
String
Force exerted by string
Water
Buoyant force
Weight
Figure 3.18
A submerged object in equilibrium
The force exerted by the string adjusts to balance the weight and the buoyant force.
Here all the forces on the submerged object are shown.
For equilibrium of this object:
force exerted by string = weight - buoyant force.
We might call the weight of the block minus the buoyant force the "effective
gravitational force".
This situation is demonstrated using a spring balance.
If the buoyant force were greater than the weight, a force would have to be applied to hold
the object down. This happens, for example with a helium-filled balloon.
Example 3.7: Floating object
Buoyant force
Air
Liquid
Figure 3.19
Weight
A floating object in equilibrium
The displaced liquid adjusts so that the forces balance.
For equilibrium, taking components in the direction vertically down:
weight - buoyant force = 0 .
The total torque will be zero if the buoyant force and the weight act along the same line.
This means that the centres of gravity of the floating object and the displaced fluids must lie in
the same vertical line.
POST-LECTURE
3-11
MOMENT OF INERTIA
The physical quantity which describes the distribution of matter about the axis of rotation is the
moment of inertia. Objects with their masses concentrated about the axis of rotation (or axle) have
smaller moments of inertia about that axis.
More precisely, if we divide the object up into small pieces each with mass Δm and at some
distance r from the axle, then the moment of inertia is
I = Σ Δm r 2
where the sum is taken over all the small pieces.
36
FE3: Equilibrium
37
Δm
Axle
r
Figure 3.20
Defining the moment of inertia
For objects with a fixed axis of rotation, total torque about the axis equals moment of inertia
about the axis times angular acceleration. This is the reason why the wheel with the metal axle in the
lecture demonstration had the larger angular acceleration.
3-12
QUESTIONS AND PROBLEMS ON EQUILIBRIUM
Q3.3 a)
A tall block sitting on a table is being pulled.
F
Figure 3.21
Tipping an object
If the pull is applied near the top, the object will tip over instead of sliding along the table. Why?
Hint: to get the block moving, the force F must be greater than the maximum frictional force acting at
the bottom on the block. F cannot be any smaller than this. Now consider torques. What point is the
block going to rotate about?
How could you pull the block along the table without tipping it?
b)
Pulling trees down with a tractor can be a dangerous occupation. Which of the methods shown
below is the less dangerous way to tie the rope to the tractor? Why?
Figure 3.22
Uprooting a tree.
Which way is safer?
Equilibrium of pivoted objects
Q3.4
Two children are balancing on a seesaw. Their weights are 200 N and 300 N . The smaller child is
1.80 m from the centre.
Suppose that the weight of the seesaw is 500 N and the centre of gravity is directly above the pivot.
Draw all the forces acting on the seesaw.
a) Calculate the magnitude of the force that the pivot exerts on the seesaw.
b) How far from the centre is the larger child at equilibrium?
Q3.5
Explain why a beam balance gives the same value for the mass of an object on the Moon as on the Earth.
3-13
STABLE, UNSTABLE AND NEUTRAL EQUILIBRIUM
We sometimes distinguish between stable and unstable equilibrium. For example, consider a cone.
FE3: Equilibrium
Stable
Stable
Figure 3.23
Unstable
Unstable
38
Neutral
Neutral
Stable and unstable equilibrium
In both cases the cone is in equilibrium because the total force is zero and the total torque is
zero. But the first case is stable, a slight displacement has no effect, while the second case is
unstable, a slight displacement causes the cone to tip over. When the cone is lying on its side it is
in neutral equilibrium.
Q3.6
What forces are acting and where are they acting in each case? What torques are responsible for restoring
the cone to its original position or otherwise?
Centre of Gravity
Q3.7 a )
b)
3-14
Does the centre of gravity always lie within an object? If not, give examples.
Suggest a way of locating the centre of gravity of a "lumpy" object (not a flat object).
FLUIDS
Pressure
When a solid object is immersed in a fluid, the force exerted on the object by the fluid is distributed
over the contact surface. For a complete description we need to look at the force acting on each
small part of the surface. We can define the average pressure on a flat surface to be the component,
δFn , of the force perpendicular to the surface divided by the area, δA, of the surface. The limit of
this quotient as we take smaller and smaller pieces of the contact surface (and hence also smaller and
smaller forces) is the pressure, P, at a point on the surface:
 δF 
... (3.2)
P = lim  n  .
δA→0 δA 
Provided that the body and the fluid are not moving, the force on each small part of the contact
surface is perpendicular to the surface (see figure 3.12) so the interaction can be described
completely in terms of pressure. (On the other hand, if there is relative motion between fluid and
solid object, the force has components parallel to the surface, not described by the pressure.)
This idea of pressure can be used also to describe what goes on inside the fluid; just imagine
the fluid divided into two parts as in the argument about buoyancy. Wherever we draw the fluid
boundary, we can define a pressure exerted by one part of the fluid on the other part. So we can say
that pressure exists within the fluid.
The following are important statements about pressure.
•
Pressure is a scalar quantity - it has no direction.
•
The pressure within a uniform stationary fluid is the same at all points in the same horizontal
plane.
•
The SI unit of pressure is the pascal, symbol Pa; 1 Pa = 1 N.m-2 .
FE3: Equilibrium
39
Density
Pressure variations within a fluid are affected by its density. The density, ρ, of a uniform substance
is defined as the quotient: mass ÷ volume:
m
ρ = V .
... (3.3)
The barometer
h
S
Figure 3.24
S'
A mercury barometer
A tube, closed at one end, is filled with mercury and is then inverted over a container of
mercury as shown. If the tube is sufficiently long, the mercury falls from the top leaving an
evacuated region there.
By considering the equilibrium of a part of the surface SS' of the mercury outside the tube, we
can show that the pressure in the mercury just under that surface is equal to the atmospheric
pressure. So too is the pressure at the same level inside the tube.
We consider the equilibrium of a body consisting of the column of mercury in the tube which
is above the level SS'. The forces on this column of mercury are its weight, W, downward and the
upward force, F, exerted by the mercury below the column. The magnitude of the upward force
must be equal to atmospheric pressure multiplied by the cross-sectional area, A, of the tube:
F = PA.
The weight of the mercury column is equal to the product of its density, its volume and g.
Since the volume is equal to the product of the column's height and cross-sectional area,
W = ρhAg .
These two forces, must have equal magnitudes, so
P = ρgh .
...(3.4)
The density of mercury is 13.6 × 1 03 kg.m-3 and normal atmospheric pressure is
1.01 × 1 05 Pa (101 kPa).
Q3.8 a ) Estimate the height of a column of mercury in a barometer.
b) The density of water is 1.00 × 103 kg.m-3.
barometer.
Estimate the height of the column of water in a water
3-15
QUESTIONS ON BUOYANT FORCES
Q3.9
Density of air
= 1.29 kg.m-3.
Density of helium = 0.18 kg.m-3.
A balloon is filled with helium. Its volume is 1.00 m3 . What downward force must be applied to stop
the balloon rising up into the sky?
Q3.10
Estimate the buoyant force exerted on you by the atmosphere.
Q3.11
An ice cube is floating in a glass of water. The water and the ice are at about 0°C.
Density of ice at 0°C
=
917 kg.m-3.
Density of water at 0°C = 1000 kg.m-3.
a) What fraction of the ice cube is submerged?
FE3: Equilibrium
40
b) Explain what happens to the level of the water as the ice melts.
Q3.12
Steel is about eight times as dense as water. How can ships float?
Q3.13
The density of water varies with temperature as shown in the graph below. The curve has a maximum at
about 4°C.
1000
Density/kg.m
-3
999
998
997
996
995
0
Figure 3.25
10
20
Temperature/°C
30
40
How the density of water changes with temperature
a) When a beaker of water is heated from below, why does the warm water at the bottom rise?
b) When a lake ices over, why does freezing occur only at the top of the lake? Is there any biological
significance in that?
Interlude 3: Masses
INTERLUDE 3 - THE RANGE OF MASSES IN THE UNIVERSE
Mass of the Universe
Mass of the Sun
Mass of the Moon
Person
Postage stamp
Protein molecule
Proton
mass/kilograms
1050 __
_
_
_
_
40
10 __ Mass of our Galaxy
_
_
_
_
30
10 __
_
_
_
Mass of the Earth
_
20
10 __
_
_
_
_
1010 __
_
_
Ocean liner
_
_
1
__
_
_
Sugar cube
_
_
-10
10 __
_
_
Red blood corpuscle
_
_
-20
10 __
_
_
_
Oxygen molecule
_
-30
10 __
Electron
41
41
FE4
MOTION OF BODIES IN FLUIDS
OBJECTIVES
Aims
By studying this chapter you will extend your knowledge and understanding of forces - in particular
the nature of forces exerted on solid bodies and small particles as they move through fluids. You
should also acquire a basic understanding of the processes of sedimentation, diffusion and
Brownian motion.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
drag force, lift force, terminal velocity, Brownian motion, diffusion, dynamic equilibrium,
sedimentation.
2.
Recognise and describe examples of situations involving forces exerted by fluids on moving
objects.
3.
Draw force diagrams and analyse forces for systems subject to lift and drag forces.
4.
Describe how the forces exerted by fluids on moving bodies vary with the speed of motion.
5.
Describe and explain how fluids affect the motion of large objects - in particular, explain how
terminal velocity is reached in certain circumstances.
6.
Describe the process of sedimentation.
7.
Explain how fluids affect the motion of small particles and the nature of Brownian motion.
8.
Describe and explain the important features of the diffusion process.
PRE-LECTURE
4-1
INTRODUCTION
The laws governing the motion of an object, as given in the previous chapters, apply only when all
the forces acting on the object are taken into account. In the real world all objects move through
fluids so we must include the forces exerted by the extremely large number of fluid molecules. In
some situations the effect of these forces can be ignored, in others they play a crucial role in
determining the motion.
We have seen in chapter FE3 that, in the special case where a large object is at rest, the forces
exerted by the fluid molecules sum to give the buoyant force, a vertical force acting through the
centre of gravity of the displaced fluid, with magnitude equal to the weight of the displaced fluid.
The forces exerted by fluids on very small objects or on moving objects are, however, much more
complicated. This lecture will give a survey of these forces and explain qualitatively how living
organisms are affected by them. In addition to its intrinsic importance, this topic is a source of
models for non-mechanical phenomena.
Questions to think about
•
When a car is driven at constant speed along a level road, a constant force is exerted on the car
by the road surface. Why doesn't the car accelerate because of this force?
•
Oxygen gas is a denser fluid than nitrogen gas. Why doesn't the atmosphere stratify into a
layer of oxygen near the earth's surface and a layer of nitrogen above it?
•
Revise the section on buoyant forces and try question 3.10 in FE3.
FE4: Motion in Fluids
42
LECTURE
4-2
FLUID FORCES ON MOVING OBJECTS
There are three important things to note about the force exerted by a fluid on a moving object which
is very much larger than a fluid molecule.
(i) The fluid force is not just the buoyant force. For example two pieces of cardboard with the
same mass and volume (and hence with the same weight and buoyant force acting on them) do not
fall at the same rate.
(ii) The force is not always vertical. For example the force exerted by the air on a ‘parachute’
moving horizontally at constant velocity must balance the other forces acting.
Hand force
Fluid force
Weight
Figure 4.1 Forces on a toy parachute
(iii) The force is not always along the line of motion. For example the fluid force on a marsupial
glider moving at constant velocity must be vertical, balancing the animal's weight.
For convenience, the fluid force is usually divided into three parts:
•
•
•
the usual buoyant force,
a lift force perpendicular to the line of motion, and
a drag force parallel to the line of motion, opposing the motion.
Figure 4.2
Forces on a glider exerted by the air
The glider's weight is not shown.
The lift and drag forces come into play only when the object is moving. Usually they increase
with speed and with the size of the object.
FE4: Motion in Fluids
43
The total fluid force does not necessarily act through the centre of gravity of the object. If it
does not, then the object may have a rotational acceleration. (See FE3.) This complication will be
ignored here.
There are some objects (such as non-rotating spheres) for which the lift force is small or zero.
Other objects produce large lift forces - aerofoils and birds' wings, for example.
There is always a drag force on an object moving in a fluid. For objects of diameter less than
about a millimetre moving in water or air, or larger objects moving in more viscous (sticky) fluids:
drag force ∝ speed .
For objects of diameter very much greater than a millimetre moving in water or air:
drag force ∝ (speed)2 .
4-3
TERMINAL VELOCITY IN A FLUID
Demonstration
A weighted balloon falls vertically through the air.
Explanation
An object falling vertically in a fluid does not accelerate at a constant rate. The downward velocity
approaches a constant value known as the terminal velocity.
Downward velocity
without drag force
Terminal velocity
with drag force
0
0
Time
Figure 4.3
Approach to terminal velocity
This behaviour is caused by the drag force. Consider the following force addition diagrams
(figure 4.4).
Total force
Total force
D
W
W
B
v =0
Figure 4.4
D
Drag force
W
B Buoyant
force
0 < v < vT
System of forces on a falling object
B
v = vT
FE4: Motion in Fluids
When the object is at rest, the forces on it are its weight and the buoyant force. The sum of
these forces (represented by the solid arrow in figure 4.4) produces a downward acceleration. As
the downward velocity increases, the drag force increases from zero, reducing the total downward
force. The downward acceleration gradually decreases to zero. The object then travels at its terminal
velocity, vT . (If there were no drag force, the total downward force and the downward acceleration
would have been constant.)
The velocity-time graph (figure 4.3) is characterised by the terminal velocity and the "rise
time" - the time taken to reach terminal velocity (or more precisely some specified fraction of the
terminal velocity). Both these quantities can be found from the equation of motion:
dv
m dt = total force.
An object travelling at terminal velocity is in equilibrium. The terminal velocity is the velocity
at which the drag force balances the other forces acting;
i.e.
total force = 0.
An approach to a constant velocity will always occur whenever there is a constant driving force
and an opposing force which increases with velocity until it is as large as the driving force.
Examples
•
Birds of prey, initially travelling faster than their terminal velocity are slowed down to terminal
velocity by the drag force.
•
Particles of different sizes and densities have different terminal velocities in water. (See the
section on sedimentation in the POST-LECTURE.)
4-4
BROWNIAN MOTION AND DIFFUSION
We now consider the motion of very small particles in fluids.
Examples
Examples include the circulation of nutrients and waste materials in living organisms. Particles
move independently of the flow of blood in animals or the flow of sap in plants. Particles of matter
move through the fluids, rather than being carried along with the fluid. What are the forces exerted
by fluids on the particles in these examples?
Explanation
Here we are looking at very small objects in a fluid which are continually undergoing random
collisions with fluid molecules. This situation is in contrast with that previously discussed, in which
the objects had diameters very much larger than 10-6 m and masses more than 1012 times the mass
of a fluid molecule. In those cases only the effects of "averaged out" forces (buoyant, lift and drag)
were detectable.
For objects smaller than 10-6 m the accelerations caused by individual collisions are larger.
Such small objects also travel further between collisions (about 10-10 m to 10-7 m) since they are
less likely to collide with fluid molecules because of their smaller sizes. The objects therefore make
a series of random movements called Brownian motion. (See figure 4.5.)
A large collection of objects, each undergoing a random Brownian motion, can move as a
whole in an organised way. This process is called diffusion.
Demonstration
A one dimensional mechanical analogue of diffusion is presented. It is shown how objects can
spread out by a sequence of purely random motions. The people move randomly using a code
based on the last digits from a page of telephone numbers: 0-3 means stand still; 4-6, move right: 79, move left.
44
FE4: Motion in Fluids
Figure 4.5
45
Brownian motion
Two important features of a diffusion process
(i) There is an overwhelming probability that the random movements will cause objects to diffuse
from regions of high concentration (of objects) to those of low concentration.
(ii) The average time for an object to move a distance d from its starting point is proportional to
d 2.
For example, glucose molecules (relative molecular mass 240) in water (relative molecular
mass 18) take 10-3 s to diffuse 10-6 m but 109 s (about 20 y) for 1 m.
Diffusion is an effective mechanism of transport of materials over typical cell dimensions (less
than about 10-5 m) in water. For example jellyfish without circulatory systems can obtain oxygen
and other nutrients from the surrounding water and lose carbon dioxide. Larger organisms require a
circulatory system and perhaps a respiratory system to bring the nutrients to the cells .
Diffusion occurs much faster in gases than in liquids because of the lower density of the fluid.
To illustrate this, the upward diffusion of bromine gas into air, is demonstrated. A state of dynamic
equilibrium is eventually reached where, at each point in space, the net rate of upward diffusion to
regions of lower bromine concentration is balanced by the rate of falling.
The final demonstration shows people participating in a two-dimensional mechanical analogue
of diffusion.
POST-LECTURE
4-5
SEDIMENTATION
Consider a small particle of density d and volume V falling with velocity v in a viscous fluid of
density ρL. The forces acting on the particle are shown in figure 4.6.
Buoyant force = ρ L Vg
Drag force = λ v
Weight = ρ Vg
Figure 4.6
Forces on a particle during sedimentation
FE4: Motion in Fluids
The constant λ in the expression for the drag force depends on the viscosity of the fluid as
well as the size and the shape of the particle.
When the particle is travelling at its terminal velocity (or sedimentation rate) vT, these three
forces are in balance.
ρ LV g + λ vT = ρ V g ;
V
vT = λ (ρ - ρL) g .
∴
Since the value of V/λ , and hence the terminal velocity or sedimentation rate, depends on the
size of the particle, sedimentation is a useful method for separating particles of different sizes. It is
used, for example, in agricultural laboratories to determine the amounts of various types of soils in a
sample.
The sedimentation rate depends on the difference in the densities of the particles and the
surrounding fluid. It is often quite slow for biological materials in water since their density is almost
equal to that of the surrounding water.
4-6
QUESTIONS
Q4.1
For a spherical particle of radius r falling in a viscous fluid characterized by a coefficient of viscosity η ,
the terminal velocity is
vT =
2r2
(ρ - ρL) g .
9η
Find the terminal velocity or sedimentation rate of a soil particle of density 2.6 × 103 kg.m-3 and
radius 0.50 × 10-6 m in water.
Viscosity of water η = 1.0 × 10-3 kg.m-l.s-l .
Density of water ρL = 1.0 × 103 kg.m-3.
Q4.2
A man with a parachute is falling vertically through the atmosphere with a velocity v . The system, man
plus parachute, experiences a drag force described by λv2 where λ is a constant.
i)
What other vertical forces are acting on the system? At terminal velocity, the vertical forces add to
give zero total force.
ii)
Find an expression for the terminal velocity in terms of these forces.
Assume that the man and parachute together weigh 850 N and that the buoyant force on them is
negligible. (See question 3.10 in FE3.)
The constant λ depends on the shape of the falling object. When the man is falling in the sky-diving
position (on his stomach with arms extended) with his parachute unopened, λ is 0.22 kg.m-l and when he
has his parachute opened up, λ is 26 kg.m-1.
iii) Find the terminal velocity of the man (and parachute) in these two positions.
If the man pulls his ripcord after sky-diving, he decelerates from the first of these terminal velocities to
the second in about 1 s .
iv) What would be the average acceleration of the man during this l s interval? (Compare this with the
acceleration due to gravity.)
Q4.3
Suppose you are driving a car along a straight level road at a constant speed of about 80 km.h-l by keeping
the accelerator pedal in a fixed position. At a high speed like this, the drag force, which is proportional to
the square of the speed, is quite large. If you depress the accelerator pedal a bit more and keep it in this
new position, the driving force on the car increases to a new constant value.
i) Explain, in terms of the drag and driving forces, what happens to the speed of the car.
ii) What happens if you then let the accelerator pedal come back to its original position? Sketch a
velocity-time graph for the complete motion.
Q4.4
In the absence of air resistance, an object thrown vertically upwards would take the same time to rise to its
maximum height as it does to return from it. With a drag force present this is no longer the case. Use
graphs of acceleration and velocity as functions of time to determine whether the object takes longer to rise
or to fall. You may neglect the buoyant force here.
(Hint: Use slopes and the areas between these graphs and the time axis where appropriate.)
46
FE4: Motion in Fluids
4-7
47
A USEFUL MATHEMATICAL MODEL
The mathematical equations used to describe the motion of a small object falling in a fluid, which
exerts a drag force proportional to the velocity, can also be used to describe currents in certain
electrical circuits and nuclei undergoing radioactive decay. Try question 4.5 if you have the
necessary mathematical expertise.
Q4.5
If the buoyant force is neglected, a small object falling vertically through the air has a downward
component of velocity, vx , at time t given by
λ

- t
mg 
vx =
1− e m  .
λ 



i)
What is the limit of the velocity (i.e. the terminal velocity) as t → ∞ ?
ii)
What is the starting velocity at t = 0 ?
iii) After what time (the rise time) is the difference between the terminal velocity and the actual velocity equal
to l/e of the terminal velocity ?
iv) Sketch the velocity component as a function of time.
v)
Show by differentiating the expression for v that the downward component of acceleration, ax , at time
t is given by
λ
t
a x = ge m .
-
vi) Sketch this acceleration component as a function of time.
vii)
Hence verify that the velocity component satisfies the equation of motion
dv
m x = mg − λ vx
dt
viii) Does the terminal velocity obtained directly from the equation of motion agree with that found in (i)?
ix) Is the terminal velocity ever reached in this model ? How does the real world differ from the model in this
respect?
x)
4-8
Show by differentiating the result of (v) that the acceleration component obeys the equation
da
m x = −λa x
dt
COLLOIDS
Insoluble particles small enough to undergo Brownian motion will not fall in a fluid. They form
what is known as a colloidal suspension which may be thought of as something intermediate
between a true suspension and a solution. Common examples of colloids include smoke particles,
the fat in homogenised milk and the protein molecules and protoplasm of cells. The particles are
kept from clumping together, usually by some kind of electrostatic repulsion. If the surrounding
fluid is altered in some way, for example by changing its acidity, temperature or chemical properties,
the colloidal particles can often aggregate and settle, as for example in the setting of jellies and
clotting of blood.
4-9
RANDOM NATURE OF DIFFUSION - QUESTIONS
Q4.6
In the two dimensional mechanical analogue of diffusion in the video lecture, you may have noticed two
people attempting to sabotage the demonstration by making precisely the same move as one another at
each stage. At each stage, each person could move in 9 distinct ways, each with equal probability.
What is the probability that two people at the same position at the same time will make precisely the
same next move, purely by chance?
What is the probability that they will make the same next ten moves?
Q4.7
Carbon dioxide gas molecules take 25 seconds to diffuse about a centimetre in air. How long would it take
them to diffuse through l metre?
Interlude 4 - Energies
INTERLUDE 4 - THE RANGE OF ENERGIES IN THE
UNIVERSE
Energy equivalent* of Sun's mass
Solar energy received by Earth annually
Earthquake, cyclone, H bomb
8 hours hard labour
KE of a rifle bullet
Energy of a nuclear bond
Ionization energy
Energy of a chemical bond
energy/joules
105 0 __
_
_
_
_
4
0
10 __
_
_
_
_
Sun's annual output
103 0 __
_
_
_
_
102 0 __ Annual world use of energy
_
_
(Energy equivalent* of 1 g of matter
_
...(Energy from fission of 1 kg of U235
_
(Burning 7000 tonnes of coal
1
0
10 __
_
_
1 kilowatt hour (3.6 × 106 J)
_
_
1
__
_
_
_
_
10-10__
Energy equivalent of a proton
_
Energy equivalent of an electron
_
_
_
-20
10 __
1 electron volt (1.6 × 10-19 J)
_
_
_
_
10-30__
* Energy equivalent of mass is calculated using E = m c 2 , where c is the speed of light in vacuum.
48
49
ENERGY
FE5
OBJECTIVES
Aims
From this chapter you will acquire a basic understanding of the concepts of work and energy. You
will not understand everything about them, because energy is an idea which has no adequate simple
definition. It is also a concept which pervades all of science, so you can expect to learn more and
more about energy throughout this course.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms:
work [mechanical work], energy, joule, power, average power, watt, kinetic energy,
conservative force, non-conservative force, potential energy, gravitational potential energy,
mechanical energy, internal force, external force, system, isolated system.
2.
State and apply formulas for work done by constant forces on objects moving along straight
paths.
3.
Calculate work and potential energy from force-displacement graphs. Sketch and plot graphs
of potential energy against position.
4.
Calculate gravitational potential energy for objects near the Earth's surface.
5.
State the conditions for the total mechanical energy of a system to be conserved.
6.
Solve simple problems involving goals 2 to 5 above.
PRE-LECTURE
5-1
INTRODUCTION
Newton's three laws of motion can, in principle, be used to study the motion of any complex system
with the following important exceptions:(i) systems of atomic dimensions or smaller, e.g. atoms in a molecule, electrons in an atom,
protons and neutrons in an atomic nucleus (here a new form of mechanics called quantum
mechanics must be used);
(ii) systems containing particles moving at speeds near the speed of light (here Einstein's
relativistic mechanics, a modification of newtonian mechanics, must be used).
Physicists do not fully understand how to treat systems of atomic dimensions moving at
speeds near that of light.
Newton's laws are a valid approximation for systems of particles which are much larger than
atoms and move at speeds small compared to that of light. Even in these cases, however, it may not
be feasible to analyse the motion using Newton's laws because of mathematical complexity
associated with large numbers of particles. Physicists use the concept of energy to treat such
systems. Energy also plays a central role in quantum mechanics.
Historically, the concept of energy was first applied to mechanical systems like colliding balls
and the pendulum. It was observed that the sum of quantities called ‘kinetic’ and ‘potential’
energies was a constant. Later the concept was generalised to include other forms - nuclear,
chemical, radiation (light), electrical, heat and sound energy. The divisions between these forms are
FE5: Energy
somewhat arbitrary. For example, thermal energy is now understood to be nothing more than
mechanical energy of the atoms and molecules of the system.
Energy is conserved but may be transformed from one form to another, e.g. mechanical energy
to heat, etc.
This chapter is one of the key parts of this course. It describes the two forms of mechanical
energy (kinetic and potential) and how energy is transferred to systems by means of a process called
mechanical work. Under certain circumstances, the total mechanical energy (the sum of the kinetic
and potential energies) of a system remains constant in time. This result can be used to study the
motions of particles in very small systems such as nuclei, atoms and molecules, as well as in large
scale systems.
The SI unit of energy and work is the joule (J); l J = 1 N.m = l kg.m2 .s-2. To give some
idea of the size of a joule: it is the energy required to raise an object weighing 1 newton (a mass of
about 0.1 kg on Earth) e.g. an apple, by 1 metre.
Power means rate of transfer of energy. In purely mechanical systems this is equal to the rate
at which work is done by some force. The SI unit of power is the watt (W); l W = l J.s-l. For
example a 100 W light globe transfers 100 J of electrical energy to heat and radiation in 1 s. And
lifting an apple weighing 1 N a height of 1 m in 1 s corresponds to an average power of 1 W.
The buoyant force exerted on objects by the air will be neglected throughout this chapter since
it is very small compared to the weight of the objects considered. (See, for example, question 3.10 in
chapter FE3.)
LECTURE
5-2
ENERGY TRANSFERS IN THE SOLAR ENERGY CYCLE
With the exception of some nuclear fuels, geothermal and tidal power, the sun is the ultimate source
of energy within the solar system. Hydrogen nuclei within the sun combine to release energy as
heat which is transferred to radiation near the surface of the sun. Some of this radiation is converted
to chemical and thermal energy at the Earth.
Radiation is absorbed directly by molecules in the atmosphere and at the Earth's surface and
appears as thermal energy. This energy is partially re-emitted as radiation; the remainder is
transferred to mechanical energy, for example in atmospheric motions and in the water cycle.
The leaves of plants use radiant energy to convert carbon dioxide and water to carbohydrates
by photosynthesis. This energy undergoes further chemical conversion in animal life and can also
be recovered by burning fossil fuels such as coal and oil. (If you are interested, see the diagram in
the interlude following this chapter and Scientific American, September 1970, for further details.)
5-3
MECHANICAL WORK - A MEANS OF ENERGY TRANSFER
Work is a measure of energy transfer. Work is done by a force acting on an object whenever the
object moves. If there are several forces acting on the same object then each of them does work.
(This is different from the colloquial use of the word ‘work’ - in the physics definition, no work is
done if there is no movement.)
Work done by a constant force for straight-line motion
Case (i): Force parallel to the motion
Consider first the simple case of a particle moving along a straight path, being acted on by a constant
force F, which acts in the same direction as the motion.
W = F (x l - x 0) ,
50
FE5: Energy
51
i.e. the work is equal to the product of the force and the displacement.
Case (ii): Force at an angle to the motion
In this case we consider a constant force in a direction which makes a constant angle θ to the
direction of the motion. Work is done only by the component, Fx , of the force which is parallel to
the line of motion. The work done by the constant force as the particle goes from position x0 to a
point x l is:
W = Fx ∆x = F cosθ (x1 - x0 )
... ( 5.1) .
•
•
•
Note the following.
Case (i) is a special instance of (ii) with θ = 0.
The component of F perpendicular to the line of motion does no work.
The work done by F is the same as that done by Fx alone.
•
The work done is positive if the component Fx is in the direction of motion and negative if Fx
is opposed to the direction of motion; i.e. if the force is ‘holding back’ the object, it does negative
work.
•
If the constant force component Fx is plotted as a function of the position x of the object, the
work done is represented by the area between the graph and the x axis from x0 to xl as shown in
figure 5.1.
Parallel force
component
Fx
W
x0
Figure 5.1
x
1
Position x
Work done by a constant force during a
straight-line displacement
Sign convention for graphs
In drawing graphs we arbitrarily choose a particular direction and call it positive. If Fx and the
displacement (∆x = xl - x0 ) are both positive or both negative, then W is positive. However if Fx
and the displacement have opposite signs (i.e. if the force is opposed to the direction of motion) then
W is negative.
Motion in a straight line with a variable force
Now consider the case where F is an arbitrary force, so that Fx varies with position x as the body
moves along a straight path. On a graph of Fx against x the work W is now represented by the area
between the curve and the x axis (with the above sign convention) See figure 5.2.
FE5: Energy
52
Parallel force component
Fx ( x )
x
+
0
-
Figure 5.2
x
1
Position x
Work done by a variable force in straight line motion
Only the component Fx of the force which is parallel to the straight path does any work.
The procedure illustrated in figure 5.2 is equivalent to the evaluating the integral
W=
x1
∫x
2
Fxdx
.
... (5.2)
General case
In the most general case of all, in which the motion is along a curved path, work can be defined by
considering a variable coordinate direction which is everywhere parallel (tangential) to the path traced
out by the moving object. The coordinate x is replaced by a coordinate measured along the path,
whose magnitude is equal to distance, s, along the curved path. Consider a graph of the tangential
component, Fˆ, of the force plotted against the coordinate s. Then the work done by the force F is
represented by the area under the curve, exactly as in the one-dimensional case.
5-4
TRANSFER OF ENERGY TO SYSTEMS BY MECHANICAL WORK
A force doing a positive amount of work on a system transfers energy to that system. The energy
can end up in one or more of the following forms:•
non-mechanical energy,
•
kinetic energy
or
•
potential energy
Non-mechanical energy
An example is a system consisting of a rigid object and a rough surface over which the object moves
(figure 5.3). The applied force does work on the object while the object moves at constant velocity,
and energy is transferred to thermal energy at the surfaces in contact. Note that there must be an
equal and opposite force - friction - acting on the object which balances the applied force, since we
are told that the object is moving at constant velocity.
Applied force
Friction
Figure 5.3 Object dragged across a rough surface
FE5: Energy
53
Kinetic energy
An example is a system consisting of an object and smooth (frictionless) surface on which it sits
(figure 5.4). When a horizontal force is applied to the object there is no opposing force - the applied
force increases the speed of the object and hence, also, its kinetic energy.
Applied force
Smooth surface
Figure 5.4 Object dragged across a perfectly smooth surface
The kinetic energy of an object of mass m travelling at speed v is defined to be
K
=
1
2
mv2 .
...(5.3)
Potential energy
An example is a system consisting of a rigid object, a smooth surface on which it sits and a spring
connected as shown in figure 5.5. Suppose the applied force is pulling the object to the right at
constant velocity, thereby stretching the spring. (Note that the applied force must be equal and
opposite to the force exerted by the spring on the object since the velocity of the object is constant.)
What happens to the energy transferred to the system as work is done by the applied force? Energy
is not being transferred to either non-mechanical energy or kinetic energy, during this constant
velocity motion.
Object
Spring
Fixed
point
Smooth surface
Figure 5.5
Spring opposing the motion
Consider a situation where the spring is stretched with the object on the right. Suppose now
that the applied force is reduced and kept slightly less than the force exerted by the spring. The
object will move to the left towards the position where the spring is unstretched. During this process
the force in the spring is doing work on the pusher (the object exerting the applied force), which is
not part of the system. If the spring is ideal, the magnitude of the work done by the spring equals
the work done by the applied force in extending the spring. The energy released by the system
comes from the potential energy (PE) stored when the spring is stretched. Potential energy should
be thought of as a property of the whole system and not of individual parts of the system.
The force exerted by the spring is called conservative since the energy transferred to the
system can be recovered directly as mechanical energy. There is always a conservative force
associated with potential energy.
Frictional forces and drag forces are called non-conservative since mechanical energy cannot
be recovered directly by removing the applied force which produced the motion.
FE5: Energy
5-5
54
CONSERVATION OF MECHANICAL ENERGY
Terminology
An internal force is one exerted by one part of a system on another part of the system. An
external force is one exerted by something outside the system on some part of the system. An
isolated system is one on which no external forces are doing work, i.e one to which no mechanical
energy is being added.
The total mechanical energy of a system is the sum of its kinetic energy and its potential
energy.
Principle
If there are no non-conservative forces acting within an isolated system, the total mechanical
energy of the system is conserved (i.e. it remains constant as time progresses).
It does not matter what we take as the zero of the potential energy because in any problem we
are interested only in changes in energy :
change in
change in
change in
total mechanical energy = kinetic energy + potential energy .
5-6
CALCULATING POTENTIAL ENERGY
How can the potential energy be found from the associated internal conservative force?
Example
In the spring example above - case (iii) - we usually choose PE = 0 when the spring has its natural
length, neither extended nor compressed. Then the potential energy at extension x is equal to the
work done by the force exerted by the spring in returning the object from x to 0.
PE(x)
Force exerted by
spring at position x
O
x
O
(a)
Figure 5.6
x
(b)
Force and potential energy for an ideal spring
There is no component of F perpendicular to the motion, so in this instance Fx is the full force
acting. The PE at x indicated in figure 5.6b is therefore represented by the shaded area in figure
5.6a.
FE5: Energy
55
General procedure
The general procedure for finding the potential energies of the various states of a system is as
follows.
(i) Assign zero PE to one state of the system.
(ii) The PE of any other state of the system equals the work done by the internal conservative
forces when the system moves from that state to the zero PE state.
5-7
WHY IS POTENTIAL ENERGY A USEFUL CONCEPT?
In general the state of a system at any instant is specified by giving the positions and the velocities
of all of the parts of the system. The PE of a state does not, however, depend on the velocities of the
particles or on how the system came to be in that state. It depends only on the positions of the
particles in the system (i.e. on their configuration). Therefore, the change in PE when the system
moves from one state to another will depend only on the configurations of the particles in the two
states and not on how the system moves between the states.
Since the gravitational force is conservative and the electrostatic and the two nuclear forces are
all conservative on atomic and nuclear scales, complicated processes can be studied comparatively
easily using PE diagrams and the principle of conservation of mechanical energy.
POST-LECTURE
5-8
ENERGY TRANSFERRED AS WORK - QUESTIONS
The following two questions are designed to help you revise the definitions given in the lecture.
Q5.1
An object of mass m slides down an incline at angle θ to the horizontal as shown.
θ
Figure 5.7
Object sliding down an incline
What are the forces on the object?
i)
On figure 5.7 draw in all the forces acting on the object.
ii) Which of these forces are doing work on the object ?
iii) Which of the forces doing work on the object are conservative ? Which are non-conservative ?
For the rest of this question assume that the non-conservative forces present are negligible.
iv) What work is done by the gravitational force on the object when the latter slides a distance d down the
incline ?
What is the increase in kinetic energy of the object during this process?
v)
F
θ
Figure 5.8
Object dragged up an incline
Draw in the forces.
FE5: Energy
56
Suppose that a force F with magnitude F greater than mg sinθ is used to pull the object a distance d up
the incline as shown in figure 5.8.
What work is done by F ?
What work is done by the gravitational force?
What is the increase in the kinetic energy of the object during this process?
Q5.2
An object of mass m swings at the end of a taut, straight unstretchable rope whose other end is fixed.
Figure 5.9
A swinging object
Draw in the forces. Which forces do work?
i)
On figure 5.9 draw in all the forces acting on the object.
ii) Which of these forces are doing work on the object?
iii) Which of the forces doing work on the object are conservative? Which are non-conservative?
Assume that the non-conservative forces present are negligible. Suppose that the object is released
from rest from some raised position with the rope taut. The work done on the object by the gravitational
force when the object moves from the point of release to the lowest point of its swing equals the kinetic
energy of the object at the lowest point. The object continues its swing until it momentarily comes to
rest at some highest point on the other side. At this highest point the kinetic energy of the object is zero.
Therefore the work done by the gravitational force when the object moves from the point of release to this
highest point is zero. (Here the object moves along a curved path so it is difficult to calculate the work
done by the gravitational force. This problem will be considered from a different viewpoint in Q5.4
below.)
5-9
GRAVITATIONAL PE NEAR THE EARTH'S SURFACE
Consider a system consisting of an object of mass m and the Earth. For our purposes the Earth may
be considered immovable so that all forces acting on it do no work. In any small region near the
Earth's surface it is possible to choose any horizontal level and say that the system has zero PE when
the object is on that level. If the object is at height h above this horizontal level the gravitational
potential energy of this system is equal to mgh. If the object is a vertical distance h below this
horizontal level, the PE is equal to -mgh . (A proof of this result will be outlined in Q5.10.)
|U|
= |mgh| ... (5.4)
PE = mgh
h
PE = 0
h
Figure 5.10
PE = -mgh
Gravitational potential energy
FE5: Energy
57
Note that these simple expressions hold only if h is small compared to the distance from the
centre of the Earth. If this condition is not satisfied, the change in the gravitational force with height
must be taken into account.
5-10
QUESTIONS
Conservation of mechanical energy
Q5.3
Consider the object sliding down the incline described in Q5.1. Assume that the non-conservative forces
acting on the object are negligible.
The system consisting of the object and the Earth is isolated according to the definition given in the
lecture. (The incline can be treated as part of the Earth.) The gravitational force on the object is now
internal to the system.
i)
Can you explain why this system is isolated?
ii) Use the idea of conservation of mechanical energy and the above expressions for the gravitational potential
energy to find the increase in kinetic energy of the object when it slides a distance d down the incline. (Do
you get the same answer as that obtained using the method of Q5.1?)
Q5.4
Consider the swinging object described in Q5.2. Assume that the non-conservative forces acting on the
object are negligible.
The system consisting of the object and the Earth (the string is not part of the system) is isolated. The
gravitational force is now internal to the system.
i)
Can you explain why this system is isolated ?
ii) Suppose that the object is released from rest from height h above the lowest point of its swing. Find the
maximum kinetic energy of the object and the maximum height to which it rises on the other side.
iii) How would your answers to part (ii) differ if non-conservative forces were not negligible? (Note that the
system would then no longer be isolated.)
Energy dissipation
Q5.5
Use energy considerations to explain why an object thrown vertically into the air takes longer to fall than
it does to rise if air resistance is taken into account (c.f. Q4.4 in FE4). Hint: consider the speed of the
object at a given height on the upward and downward portions of its motion.
Energy balances
Q5.6
Flow of sap in trees
A typical tall tree manages to raise water through a height of 20 m from roots to leaves at a rate of
700 kg per day. (Approximately 90% of this water is evaporated from the leaves, the remainder is used in
the photosynthesis process.) What energy must be supplied in a day to raise this water?
It is impossible to raise water to a height of more than about 10.3 metres using a vacuum pump which
utilises atmospheric pressure. How do trees manage to accomplish this? (See Scientific American, March
1963, pp 132-142.)
Q5.7
v
Figure 5.11
At what rate does the possum work?
The metabolism of a possum of mass m is such that the possum is capable of doing work at a
maximum rate of P. When the possum climbs a tall vertical tree at constant speed, it must do work at a
rate equal to the rate of increase of the potential energy of the possum-Earth system. (We can neglect any
non-conservative forces for the purposes of this discussion.)
i)
If the possum is moving at speed v, at what rate is it doing work?
ii) If m = 3.0 kg and P = 60 W what is the shortest possible time in which the possum can climb a tree 20
m high ? (These ideas will be used in chapter FE8 on Scale)
FE5: Energy
58
Machines
Machines usually are devices for doing work using a small force travelling a large distance. For
example on an inclined plane like a winding road, only a force large enough to overcome the
component of gravity parallel to the plane (and any non-conservative forces present) is required.
However, one must travel a longer distance to reach any given height.
Q5.8
When using a lever like a crowbar, one applies a force at one end to raise a load at the other (figure 5.12).
a
b
Applied
force
Load
Figure 5.12
Lifting with a lever
Compare forces and works.
i)
By considering torques about the pivot, find the ratio of the load, mg, to the minimum applied force F
required to raise the load, in terms of the ratio of the lengths a and b of the lever on either side of the pivot
point. You may neglect the weight of the lever.
ii) If the load is raised a small height h, the work done by the minimum applied force must equal the
increase, mgh, in the PE of the load-Earth system. How far does the applied force have to move its end of
the lever? (Ignore the fact that the ends of the lever move in slightly curved paths.)
Q5.9
Draw a schematic diagram of the main bones in the arm and the biceps muscle used to raise the forearm.
Discuss qualitatively how this ‘machine’ in the human body differs from the machine used in Q5.8. Hint:
consider the work done by the biceps muscle in raising a load held in the hand.
Gravitational force and potential energy
In questions 5.3 to 5.9 you used the result that the gravitational potential energy of an object-Earth
system depends only on the height of the object. Question 5.10 outlines a proof of this result.
Q5.10
B
A
E
D
C
F
Figure 5.13
Work done by gravity
Show that the work done by gravity does not depend on the path taken.
In figure 5.13 A is on the same horizontal level as B which is vertically above C , a distance h below.
i)
Find the work done by the gravitational force on an object of mass m when the object moves from A to B
as shown.
ii) Find the work done when the object moves from B to C and hence the total work done when the object
moves from A to C along the path A → B → C.
iii) The path A → D → E → F → C is made up of straight vertical and horizontal segments as shown.
Verify that the total work done by the gravitational force on the object when it moves from A to C along
this path is equal to the result of (ii).
FE5: Energy
59
A
C
Figure 5.14
Straight-line approximation to a curved path
Any curved path from A to C can be approximated as closely as desired by a series of vertical and
horizontal segments - see figure 5.14. The work done along the curved path is then approximately equal to
the total work done along the straight line segments, i.e. the result of (ii) .
Thus we can unambiguously define the difference in PE between the states of the object-Earth system
represented by A and C to be the work done by the gravitational force when the object moves from A to C
along any path, i.e. PE is equal to mgh in every case.
Other conservative forces and potential energy
Q5.11
Wall
Moving
object
x
Figure 5.15
Object striking a spring buffer
What happens to its kinetic energy?
Figure 5.15 shows a spring-buffer arrangement to prevent a moving object from striking a wall. The force
F exerted by the spring-buffer on the object is plotted in figure 5.16 as a function of the distance x of the
object from the wall.
F / kilonewton
8
4
0
G
0
Figure 5.16
E
D
C
1
B
A
2
3
x / metre
Force exerted by the buffer spring on the block
Choose the PE of the system to be zero when the moving object is 3 metres from the wall, i.e. at the
point labelled A .
i)
Find the potential energies at points B, C, D, E and G by calculating the work done by F when the object
moves from those points to A . (Hint: Use areas under the graph.)
ii) Sketch the potential energy diagram for the system.
iii) Use the diagram to find the minimum kinetic energy of the object for it to strike the wall.
iv) How would the answers to (i) - (iii) be altered if the PE of the system had been chosen to be zero at
x = 0?
FE5: Energy
60
Example: Atoms in a diatomic molecule
The force between atoms is a complicated one due to individual electrical forces exerted by the
nucleus and the electrons of one atom on those of the other atom. There is an infinite repulsive force
at zero separation and the force decreases and becomes attractive at larger separations (figure 5.17).
Repulsive force
Component of
force exerted
by atom at
0 on atom
at position
x
x
0
Separation
Attractive force
Figure 5.17 Force between atoms
The graph shows the force exerted by one atom on the other. The component of the force is taken in
the direction of the line joining the atoms. Positive values of the component correspond to repulsion;
negative values represent attraction
The zero of PE for the two atoms is conventionally defined to occur when the atoms are a very
large (infinite) distance apart. Can you see why it would not be sensible to define PE = 0 when x =
0?
The PE is then the work done by F(x) when the separation is increased from x to ∞ , i.e. the
shaded area in figure 5.17. Although the curve extends to very large distances (∞) this shaded area
is finite.
PE(x)
0
x
0
Separation
C
A
B
Figure 5.18
Potential energy for the system of two atoms
By convention the PE is taken to be zero when the separation is infinite.
Q5.12
Use the section, Sign convention for graphs, on page 51 to explain why the PE is negative at the
point marked A in figure 5.18 .
At point B , the PE is minimum. To which point on the force-separation curve does this correspond?
How does the force behave on either side of this point ? What happens to a stationary object placed at that
point? At point C, the PE is zero. To which point on the force-separation curve does this correspond?
FE5: Energy
5-11
61
FINDING THE CONSERVATIVE FORCE FROM THE PE CURVE
Questions 5.10, 5.11 and 5.12 were practice in finding the PE curve for a system from the
appropriate conservative force. It is worth noting in passing that since we obtain the PE curve from
the conservative force by integration (areas under curves) we may also find the force by
differentiating (finding the slope) of the PE curve (c.f. velocity and distance).
d Conservative force component = - dx (potential energy)
=
- (slope of PE vs x curve).
The minus sign is needed to give the correct direction for the conservative force.
This result can be used to give a quick check on whether the PE curve obtained from a
conservative force is correct : the force should be negative if the PE increases as x increases, positive
if the PE decreases and zero if the PE is constant. Try this out on the PE curves you have met so
far.
5-12
CONCEPTUAL MODELS FOR POTENTIAL ENERGY
If you are finding PE curves to be rather abstract, you may like to think of them as depicting a
succession of hills and valleys viewed from the side (see figure 5.19).
PE(x)
PE(x2 )
PE( x 1 )
0
x1
Figure 5.19
x
2
x
Potential energy hills and valleys
The difference between the potential energies at x1 and x2 is just like the difference between
the gravitational potential energies when the object is at different heights on the hills above x1 and x2 .
Consider an object sliding without friction up and down the ‘hills’. When viewed from
directly above, only the horizontal motion in a straight line can be seen. The object appears to speed
up and slow down so there must be a horizontal force acting on it. This force at any point is
obviously related to the slope of the hill at that point (c.f. previous section). The PE surface seen in
the TV lecture was a generalisation of the PE curve to a case where the object can undergo a two
dimensional motion.
5-13
POWER
Power, the rate at which energy is supplied to, released by, or dissipated within a system is often a
quantity of interest.
The average power during a time interval is the energy transferred during that time interval
divided by the time interval.
Q5.13
Calculate the average power required to raise the water in the tree of Q5.6.
You will meet energy again in all of the other units of this course. The unit Thermal Physics in
particular is concerned mainly with transfers of energy to and from systems.
62
INTERLUDE 5 - EARTH'S ENERGY BALANCE AND FLOW
Note:
1 TW = 1 × 101 2 W.
Solar radiation
178 TW
Short wavelength
radiation
Long wavelength
radiation
Direct reflection
62 000 TW (35%)
Tidal
energy
Tides, tidal
currents etc.
3 TW
Direct conversion
to heat
76 000 TW (43%)
Evaporation, precipitation,
runoff etc.
40 000 TW (22%)
Storage:
water & ice
Convection,volcanoes,
hot springs
0.3 TW
Winds, waves, convection and curre nts
Photosynthesis
40 TW
Storage:
plants
Decay
Conduction
32 TW
Animals
Terrestrial enegy
Fossil fuels
Nuclear, thermal &
gravitational enrgy
63
FE6
ROTATION
OBJECTIVES
Aims
A careful study of this chapter should equip you with a basic understanding of the concepts and
principles used to describe rotational motion. You will learn how to apply these ideas to fairly
simple cases where things rotate about an axis whose direction in space remains fixed. You should
understand the idea of a frame of reference and the kinds of modifications to physical theory which
are adopted in order to cope with situations observed from an accelerating frame of reference.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
angular velocity, angular acceleration, centripetal force, translational kinetic energy,
rotational kinetic energy, frame of reference, pseudoforce, centrifugal force, centrifuge,
ultracentrifuge.
2.
Solve simple problems on uniform circular motion involving centripetal force, centripetal
acceleration, mass, radius, speed and angular velocity.
3.
Explain how the total kinetic energy of a rigid body can be expressed as the sum of
translational and rotational kinetic energies and apply this result to simple problems.
4.
Explain why and how the equation of motion for an accelerating frame of reference is
modified by including pseudoforces
5.
State and apply the formula for centrifugal force in terms of mass, angular velocity and radius.
6.
Explain the operating principles of the centrifuge and ultracentrifuge.
PRE-LECTURE
The theme of this chapter is rotation. We shall examine two quite different aspects. Firstly,
extending the discussion on energy in the previous chapter, we shall look at the kinetic energy of
rotating objects. Secondly, we shall look at the centrifuge. In order to talk about the centrifuge, we
digress and see how the equation of motion may be written in different frames of reference; in
particular, how it may be written in a rotating frame of reference.
FE6: Rotation
6-1
64
CIRCULAR MOTION
Consider a small object moving with constant speed v in a circle of radius R.
v
v
dθ
R
Figure 6.1
Circular motion
In a short time interval dt it sweeps out an angle dθ . The angular velocity of the object is
defined to be
dθ
ω = dt .
... (6.1)
The angular velocity and the speed of the object are connected by the relation
v
ω = R .
... (6.2)
If the angular velocity changes with time the object also has an angular acceleration:
dω
α = dt .
... (6.3)
The SI unit of angular velocity is the radian per second, symbol rad.s-l. The unit of angular
acceleration is radian per second per second, rad.s-2.
6-2
ROTATION OF A RIGID BODY ABOUT A FIXED AXIS
The reason for introducing these angular variables is to simplify the description of rotating bodies.
If a rigid body rotates about an axis which is fixed in space, then different parts of the body have
different linear velocities and accelerations. On the other hand, all the particles in the body have the
same angular velocity and the same angular acceleration. This description in terms of angular
variables is useful even if the axis of rotation moves.
Q6.1
In chapter FE1 we quoted the result that the centripetal acceleration of an object moving with speed v in a
v2
circular path was equal to
. Re-express this in terms of angular velocity. (We need this result later.)
R
Preparation
Read the sections in chapter FE3 about rotational motion and moments of inertia. Revise questions
2.4 and 2.6(b) in FE2.
FE6: Rotation
65
LECTURE
6-3
ROTATIONAL KINETIC ENERGY
When a rigid body is rotating about a fixed axis, different parts of the object move with different
speeds. Parts near the edge have greater speeds than those near the axis of rotation. Consider one
1
small part of the object, with mass Δm, moving at speed v. Its kinetic energy is 2 (Δm) v2 , so the
total kinetic energy of the whole body is the sum of the kinetic energies of all its parts, which we can
1
write as Σ 2 (Δm) v2 . Note that there will be many different values of v in this sum. However, by
using the common angular velocity, ω, instead of speed we can write a simple formula for the
rotational kinetic energy:
KE
=
1
2
Iω
2
... (6.4)
where I is the moment of inertia of the object, mentioned in chapter FE3. The moment of inertia
depends on the distribution of mass in the body. If you consider two objects with the same total
mass, the one with more of its mass further from the axis of rotation has the higher moment of
inertia. Note also that a body does not have a unique moment of inertia; the value of I depends on
the location of the axis of rotation.
In general a rigid body (e.g. a boomerang) has both rotational and translational motion. Each
1
part still has kinetic energy 2 (Δm) v2 , but some of that energy is now associated with the motion of
the body as a whole and the rest of it is associated with the rotational motion. Consequently the total
kinetic energy of the body can be expressed as a sum of translational kinetic energy, associated
with the motion of the centre of gravity of the body, and kinetic energy of rotation. Provided that the
body is rigid and that its axis of rotation through the centre of gravity stays in the same direction in
space, the total KE can be written in the simple form:
total KE
=
1
2
m vC2
1
+ 2
Iω 2 ,
where vC is the speed of the centre of gravity and m is the total mass of the body.
Demonstrations
• A block of dry ice slides down a slope more quickly than a round object rolls down the slope.
Gravitational force does work on the objects. All the PE of the sliding object is converted into
KE of translation. But the PE of the rolling object must be shared between KE of translation
and KE of rotation which means that the translational motion takes only a fraction of the total
KE. (The value of that fraction depends on the shape of the body, but not its size.) At a given
distance down the slope, the speed of the centre of gravity must be less for the rolling object.
• A sphere rolls more quickly than a solid cylinder which in turn rolls more quickly than a
hollow cylinder, even though all three objects have equal radii. This happens because the mass
is distributed differently in each object and the sharing between the KE of translation and the
KE of rotation is different.
• In a model car energy is stored as rotational KE of a flywheel. It is converted into translational
KE of the car and eventually into non-mechanical energy which is lost from the system.
FE6: Rotation
6-4
66
ACCELERATED FRAMES OF REFERENCE AND PSEUDOFORCES
Frames of reference
If you want to describe the motion of an object you need a coordinate system, or something
equivalent, to which you can relate your measurements of position; it doesn't make sense to give
position coordinates (x, y, z) unless you know where the origin is and the directions of the axes.
Specifying the origin and the axes is a formal, but somewhat abstract, way of describing a frame of
reference. In practice a frame of reference is anchored to some physical objects, and when we talk
about an observer we imagine a person at rest in his or her own reference frame. We now explore
some aspects of how the choice of a reference frame can affect the observed motions of objects and
even the laws of motion themselves.
Demonstration: Acceleration in a straight line
•
A truck is moving at constant velocity (figure 6.2). An object is dropped and is seen to fall to
a point on the floor of the truck, directly below the point of release. To an observer in the truck's
frame of reference, the object falls vertically along a straight line path. The truck's frame of reference
is not accelerating, and nothing unusual occurs.
Truck moves at constant velocity
Object released here ...
No Acceleration
... strikes floor here
Figure 6.2
Observer moving with truck
Dropping an object in a moving frame of reference
The frame of reference attached to the truck is moving with constant velocity.
•
In the second part of the demonstration the object is dropped while the truck is slowing down
to stop (figure 6.3).
Figure 6.3
Dropping an object in an accelerated frame
In this case the truck and its frame of reference are accelerating. To an observer in the truck's
frame of reference it looks as if there is a force pushing the object forward as it falls. The observer
cannot explain this using the equation of motion. There is no known kind of forward force to
FE6: Rotation
67
account for the path of the falling object. The problem can be resolved by introducing a fictitious
force, or pseudoforce, to make the equation of motion work in the accelerated frame of reference.
Demonstration: Measurement of weight in a lift
A body is weighed in a lift. (This problem was described in Q2.10, FE2, where a fixed frame of
reference, outside the lift, was used.) If the frame of reference is attached to the accelerating lift, then
the forces on the body being weighed are the gravitational force, the supporting force exerted by the
platform of the scales, plus a pseudoforce, associated with the accelerated frame (figure 2.11). The
supporting force, whose magnitude is equal to the apparent weight recorded on the scales depends
on both the gravitational force and the pseudoforce.
Motion in a circle
Imagine an object which has a string tied to it and is swung around in an approximately horizontal
circular path (figure 6.4). Think of an imaginary observer riding on the object. The frame of
reference for this observer is accelerating. Now the observer knows that the string exerts a force to
the left, but cannot explain why the object does not go off in that direction, although the equation of
motion predicts that it should. To make the equation of motion work the observer introduces a
pseudoforce - in this case a force to the right in order to counteract the force exerted by the string.
?
Observer moving with
the object
Force
exerted by
string
Figure 6.4 Observer in a rotating frame of reference
We know from our bird's eye view of the object moving in the circle that the force exerted by
the string is equal to mω 2 R. Therefore the moving observer's pseudoforce must also be equal to
m ω 2 R, but it is in the opposite direction - outwards. This kind of pseudoforce is often called
centrifugal force.
Force
exerted by
string
Figure 6.5
Pseudoforce
m ω 2R
Forces in a rotating frame
In summary, the equation of motion works properly only in a frame of reference that is not
accelerating. However it is often convenient, from the point of view of mathematical simplicity, to
use an accelerated reference frame and non-physical pseudoforces which are invented only in order
to preserve the equation of motion. We say that the pseudoforces are non-physical firstly because
they violate the law that forces always occur in pairs and secondly because it is not possible to
identify a physical object which is the source of the force.
FE6: Rotation
6-5
68
THE CENTRIFUGE
The centrifuge is used extensively to separate materials of differing densities. A tube containing
the materials swings around the axis of the centrifuge at a high angular velocity (figure 6.6).
Axis
Figure 6.6
Principle of the centrifuge
If a suspended particle is going to remain at a given radius as the centrifuge rotates, the forces
on it due to the surrounding material must combine to provide a centripetal force m ω 2 R. If the
surrounding material cannot provide sufficient force the particle moves to a greater radius where the
surrounding material or the bottom of test-tube can provide the necessary centripetal force. (See
figure 6.7.)
Figure 6.7
Path of a particle in a centrifuge
This is a bird's eye view of the path. The particle moves to greater radius
Our discussion of the centrifuge can be greatly simplified by looking at things from the point
of view of an imaginary observer who rides around with the test tube rather than from a bird's eye
view. An observer riding around with the tube in the centrifuge is accelerating so we can use the
equation of motion (F = ma) only by introducing a fictitious centrifugal force.
Consider the motion of a suspended particle as described in the accelerated frame of reference.
In figure 6.8 the force to the left is provided by the surrounding material.
Axis of rotation
ROTATING FRAME
Contact force
Centrifugal force
R
Figure 6.8
Real force and pseudoforce on a particle in a centrifuge
In this case the particle is suspended at a fixed radius, so it is at rest in the rotating frame. The
centrifugal force is a pseudoforce. The buoyant-like contact force is a real pressure force exerted by the
surrounding fluid.
FE6: Rotation
69
The pseudoforce is still equal to mω 2 R. If the real contact force provided by the surrounding
material is less than or equal to m ω 2 R then the particle moves further down the test-tube. The
centrifugal pseudoforce is rather like a gravitational force which causes suspended particles to settle
out. By making the angular velocity (ω) very high the pseudoforce (mω 2 R) can be 104 to 105 times
greater than the weight of the particle. So we can ignore the weight, giving a total force along the
radius. Seen from the rotating frame, the motion of a particle that starts from rest will therefore be a
straight line, not the spiral seen by the outside observer.
6-6
THE ULTRACENTRIFUGE
The ultracentrifuge in the Department of Biochemistry is often used to measure the molecular
weights of protein molecules. The rotor shown in the lecture runs at speeds up to 60 000
revolutions per minute giving a maximum centrifugal force of 260 000 times that of gravity.
Rotors for ultracentrifuges must be carefully designed so that they do not fly apart under the large
centrifugal forces. These considerations set a limit on the radius of a rotor and its speed of
operation. The chamber containing the rotor is evacuated to eliminate heating due to air friction.
Cell
Cell
Figure 6.9
An ultracentrifuge rotor
A cut-away view showing the location of cells containing protein samples.
In this case, the particles of interest are very small - small enough for diffusion to occur; so
they are also small enough to be influenced by the forces exerted by individual liquid molecules
rather than by the ‘averaged out’ forces (the drag and the buoyant forces). Sedimentation of the
protein molecules cannot take place under gravity because their weight is not sufficient to overcome
their diffusion upwards. If we centrifuge the sample, we can increase the speed of rotation until the
centrifugal (pseudo) force overcomes the diffusion and drives the protein molecule to a greater
radius.
The molecular weight of the protein molecules is determined from the sedimentation rate, the
density of the molecules and the density of the surrounding fluid. Diffusion effects must be
subtracted out. The size of the diffusion effects is obtained from measurements of the
sedimentation rate taken at different speeds of rotation.
The measurements are all made while the sample is spinning around inside the machine. This
is the difference between an analytical ultracentrifuge and the ordinary type of centrifuge which is
used for separation. The radial movement of the protein molecules is monitored by optical
techniques. Use is made of the fact that the refractive index of the protein molecules is different
from that of the surrounding liquid.
6-7
CORIOLIS FORCE
Another kind of pseudoforce, the Coriolis force, appears in a rotating frame of reference. A
demonstration shows a two-dimensional version of the Coriolis effect. A container moving along a
straight path drops sand onto a rotating platform. The track of the falling sand in the frame of the
FE6: Rotation
70
platform is a curve rather than a straight line. To explain this curved path another fictitious force, the
Coriolis force, can be introduced.
A similar effect occurs with global wind patterns. Air masses moving towards the equator
are deflected from a true south-north path because the Earth rotates underneath. In a rotating frame
attached to the Earth's surface, the winds appear to be deflected. In order to preserve the equation of
motion in the rotating frame we invent two fictitious forces - centrifugal and Coriolis.
POST-LECTURE
6-8
QUESTIONS
Rotation
Q6.2
A gramophone record is rotating at a rate of 33 l /3 revolutions per minute. What is its angular velocity in
rad.s-l?
Q6.3
The dental drill probably reaches higher speed than any other production micro-mechanism. The angular
speed is about 60 000 rad.s-l. If the dental burr has a diameter of 1.5 mm, how fast is the outside of the
burr moving?
Rotational kinetic energy
Q6.4
A flywheel is rotating with an angular velocity of 5 revolutions per second. What does its angular
velocity become when its rotational kinetic energy increases 16 times?
Q6.5
Could a sphere roll down a slope if friction were absent?
Q6.6
How can you tell the difference between a raw egg and a hard-boiled egg by spinning them?
Accelerated frames of reference and pseudoforces
Q6.7
Refer to the demonstrations of motion observed in a frame of reference attached to a moving truck.
a)
If the truck were speeding up (figure 6.10) where would the object strike the floor?
Object released here
Truck is speeding up
Acceleration
Figure 6.10
An accelerating frame
What happens to the falling object?
b) Can an observer standing at the side of the road explain the motion using the equation of motion? (Or
does the observer need to invoke pseudoforces?)
Q6.8
6-9
Discuss the way you are thrown around in a car when it stops suddenly and when it turns a corner sharply
i)
from the point of view of a bird watching from its perch on a power line, and
ii)
from the point of view where all measurements are made with respect to the car.
MORE ABOUT THE CENTRIFUGE
The centrifuge is used to increase the settling speed of particles which fall very slowly through the
surrounding liquid. (See §4-5 on Sedimentation). In a frame of reference rotating with the test-tube
the horizontal forces on a typical particle are as shown in the horizontal plan view in figure 6-11. In
comparison with these forces, the vertical forces on the particle are very small and may be neglected.
(The Coriolis force can also be neglected in this argument.) The two forces acting to the left are
exerted by the surrounding liquid. The drag force is the same as that introduced in FE4. The
FE6: Rotation
71
‘buoyant-like’ force is the analogue of the familiar buoyant force - its magnitude is independent of
the motion of the particle relative to the liquid. Both are real forces.
Figure 6.11
Real forces and pseudoforce on a particle in a centrifuge
Here the particle is moving ‘down’ the tube to the right. The ‘buoyant like’ contact force and the drag
force, both exerted by the fluid, are real forces. The centrifugal force is a pseudoforce.
The magnitude of the buoyant-like force can be found by the method used for the buoyant
force in FE3. Consider the forces which would act on a portion of fluid, mass mL, corresponding to
the liquid displaced by the particle. See the upper part of figure 6.11. Since this portion of liquid
does not move along the radius (why?), the ‘buoyant like’ force balances the centrifugal force mL
ω 2 R. Using ρL to denote the density of the liquid and V for the common volume of the particle and
the displaced liquid we get:
buoyant-like force = mL ω 2 R = ρLVω 2 R.
Returning to the forces on the suspended solid particle (lower diagram in figure 6.11), the
buoyant like force on the particle is equal to that on the displaced liquid. (Why?) Also
centrifugal force on the particle
where ρ is the density of the particle.
=
ρVω 2 R ,
The terminal velocity is the velocity at which the drag force λv balances the other forces acting.
So the equation of motion for the suspended particle becomes:
ρ LV ω 2R + λvT = ρVω 2 R.
We can rearrange this to get an expression for the settling speed:
V
v = λ (ρ - ρL) ω 2R.
Compare this with the equation for gravitational settling worked out in §4-5 :
V
vT = λ (ρ - ρL) g .
2
The angular speed ω of the centrifuge can be increased to make the centripetal acceleration, ω
R , very much greater than g and to provide very much greater settling speeds.
Q6.9
A general laboratory centrifuge rotates at its maximum angular speed of 8 000 revolutions per minute.
Consider a particle at a rotation radius of 110 mm.
a)
b)
Compare the centrifugal force on the particle with its weight.
In Q4.l, FE4, the sedimentation rate for a particle settling under gravity was several millimetres per
hour. What would the sedimentation rate be if this centrifuge were used?
72
SUMMARY:
GRAPHICAL PRESENTATION OF INFORMATION
The table below shows how slopes and areas of various kinds of graphs can be used to calculate
values of physical quantities.
Ordinate (‘y’)
Abscissa (‘x ’)
Slope
Area between curve
and ‘x’ axis
distance travelled
time
speed
--------
displacement*
time
velocity*
--------
speed
time
---------
distance travelled
velocity*
time
acceleration*
displacement*
acceleration*
time
---------
change in velocity* (v)
total force*
time
----------
mass × (change in
velocity*)
force*
displacement*
---------
work
conservative force*
displacement*
---------
- (change in potential
energy)
potential energy
displacement*
-(conservative force*)
------------
73
FE7
OSCILLATIONS
OBJECTIVES
Aims
By studying this chapter you can expect to understand the nature and causes of oscillations. You
will need to learn a fair number of new terms, but some care and effort in doing that will be well
rewarded later because the ideas and principles introduced here can be used to understand a wide
range of natural phenomena. In particular, the concepts and language of oscillations form the basis
for understanding waves and optics, which come later in this course.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
oscillation, simple harmonic motion [SHM], amplitude, period, frequency, angular
frequency, phase, initial phase, free oscillation, damped oscillation, forced oscillation,
natural frequency, restoring force, force constant [spring constant], driving force, damping
force, transient motion, steady state oscillation, resonance, resonance peak, resonant
frequency.
2.
Explain free, damped and forced oscillations in terms of forces and energy transfers.
3.
State and apply the equations and formulae of one-dimensional SHM for
(a)
displacement, velocity and acceleration in terms of time;
(b)
restoring force and acceleration in terms of displacement;
(c) frequency and angular frequency in terms of force constant and mass for an object on a
spring.
4.
Describe the important properties of SHM and explain how SHMs can be combined to give
more general oscillations.
5.
Describe and explain the phenomenon of resonance and recognise examples of resonance.
PRE-LECTURE
7-1
SIMPLE HARMONIC MOTION (SHM)Oscillations (periodic to and fro motions) of
objects occur frequently both in nature and technology. For example, sound waves, water waves,
elastic waves in stretched strings etc. are made up of large numbers of objects oscillating in a
coordinated manner. This chapter will describe the various types of oscillations in terms of the
forces and energy transfers necessary for their existence.
Consider a particle moving in one dimension so that its displacement from a fixed point at time
t is given by
x = A cos(ω t)
where A and ω are constants. This equation describes a simple harmonic motion. The following
questions are designed to lead you through some important characteristics of SHM.
FE7: Oscillations
74
Q7.1 i) Where is the particle at t = 0 ?
ii)
Sketch a displacement-time graph of the motion. Note that the graph would look the same if you shifted
2π
it either to the right or to the left by a time interval T =
.
ω iii) What is the largest value of |x | ?
At what times is the distance from the centre, i.e. |x|, largest?
iv) Verify that the x-component of velocity, vx =
v)
dx , at time t is equal to - A ω sin(ω t) .
d t Hence verify that the speed is smallest when |x| is largest and largest when |x| is smallest. (Do these
results agree with those obtained directly from the slopes of the displacement-time graph?)
vi) Verify that the acceleration component, ax =
dvx
dt
, at time t is -Α ω2 cos( ω t).
•
The acceleration is proportional to the displacement x and is always directed towards the
centre, x = 0. The proportionality constant is -ω2 ; that is, the acceleration component is given by:
ax = -ω2 x
vii)
...(7.1)
Show that the displacement, velocity and acceleration will all have the same values again after a further
2π
time T =
.
ω •
The motion is cyclic, continually repeating itself.
As you will have seen in (i), at time t = 0, the displacement is A ; the particle is at one of the
extreme points of the oscillation. It is often useful to start the description of the oscillation at some
other part of the cycle. Mathematically this corresponds to writing
x = A cos(ωt + φ )
... (7.2)
φ
= A cos[ω (t +
)]
ω or shifting the displacement-time graph a distance Δt = φ / ω to the left. Since the velocity
component at any time t is just the slope of the displacement-time graph, and the acceleration
component is the slope of the velocity-time graph, the velocity-time and acceleration-time graphs are
also shifted to the left by the same amount, φ /ω. The relationships between these three quantities, x
, vx , and ax are unaltered by the shift. (The same initial phase φ appears in each.)
Equation 7.2 is the most general expression for the displacement of a particle undergoing
simple harmonic motion. As the name suggests, it is the simplest form of oscillatory motion; its
importance lies in the fact that more complex oscillations can be studied in terms of it.
Terminology
Symbol
Name
SI unit
amplitude
m
angular frequency
s-l
(or rad.s-l)
phase
no unit
(or rad)
φ
initial phase (i.e. phase at t = 0)
no unit
(or rad)
T
period
s
f
frequency
Hz
A
ω
ωt +φ
FE7: Oscillations
75
Frequency, angular frequency and period for any kind of oscillation are related as follows:
ω
1
f = T = 2π
... (7.3).
The SI unit for frequency is the hertz (Hz); 1 Hz = 1 cycle per second. Phase is a dimensionless
quantity, so it needs no unit. However, since it is often helpful to visualise phases as angles, they are
often given the unit radian. Note that there are 2π radians in each cycle, so a frequency of 1 Hz
corresponds to an angular frequency of 2π s-l or 2π rad.s-l.
Two oscillations of the same frequency are said to be in phase at any given time if they have
the same phase at that time. Otherwise they are out of phase by the difference of their phases
(φ2 - φ1).
In phase
Out of
phase by
π/2
Figure 7.1
Oscillations in phase and out of phase
LECTURE
7-2
FREE OSCILLATIONS
If the oscillating system is isolated (i.e. if no energy is being added to or taken away from the
system) the oscillations are called free oscillations. Oscillations (free or otherwise) can occur
whenever a restoring force capable of transforming potential energy (PE) to kinetic energy (KE) and
vice versa is present. In a free oscillation, since the sum of the PE and KE cannot increase, the PE
must be largest at the extreme points of the oscillation where the KE is zero.
Examples
•
Liquid sloshing mode - the restoring forces are due to gravity.
•
A vibrating metal plate - elastic restoring forces.
•
Stretched string - the restoring force is provided by tension in the string.
In each of these three examples all the oscillating particles together formed a standing wave
pattern.
FE7: Oscillations
76
Example of a single oscillating object
A ball rolls in a curved track. When viewed from above, the ball is executing a one-dimensional
oscillation. The restoring force for the horizontal oscillation is the horizontal component of the force
exerted on the ball by the track. (The ball also oscillates vertically.)
Contact force
C
CV
Torque
Torque
CH
Weight
W
W
Figure 7.2
A ball oscillating on a curved track
The ball is acted on by two forces (left), a contact force C and the gravitational force W. The force C
produces a torque about the centre of the ball because its line of action does not pass through the centre.
An equivalent representation (right ) shows the single contact force replaced by a vertical contact force
and a horizontal contact force.
Since the track can have any shape (provided that every point in the middle is lower than the
ends) there are infinitely many possible restoring force functions and therefore there are infinitely
many different one-dimensional oscillations.
SHM - the simplest free oscillation
Any oscillation can be represented as a combination of SHMs for the following reasons.
•
Any one-dimensional oscillation (free or otherwise) can be represented mathematically as a
combination of SHMs with appropriate frequencies, amplitudes and initial phases. An example is
the synthesis of the displacement-time graph of a sawtooth oscillation from SHMs.
•
Any two- or three-dimensional oscillation can be treated as two or three one-dimensional
oscillations at right angles to one another, since the motions in those directions are independent.
(Some examples, Lissajous figures in two dimensions, are shown in §7-7, figure 7.10.)
When an object is undergoing SHM, its acceleration and the total restoring force at any time
are both proportional to the displacement from the centre and they are both directed towards the
centre.
Example: Body on a spring
Consider a body on the end of a spring, moving on a frictionless surface.
FE7: Oscillations
Fixed
point
77
F
x
Figure 7.3
A system which does SHM
For an ideal spring, the x-component F of the restoring force is equal to -k x , where k is a
constant; i.e. F is proportional to displacement and is directed towards the equilibrium point where x
= 0.
k
The equation of motion is
max = −kx or
ax = − m x .
Comparison with equation 7.1 gives the natural frequency of oscillation:
k
1 k
ωN = m
or
fN = 2π m .
... (7.4).
Energy
In a free oscillation, since the only force doing work is conservative; the total mechanical energy (KE
+ PE) of the system is a constant.
PE(x)
KE = 0
Total
mechanical
energy
KE
PE
0
-A
Figure 7.4
x
A
Energy of a free oscillation
The curve shows the potential energy (PE). The line across the top represents the level of the constant
mechanical energy which is also the maximum value of PE. As PE increases, KE decreases. When
displacement is an extreme value, x = A or x = -A, KE is zero and PE is at its maximum. For SHM
the curve is a parabola; for other oscillations the shape is different but the energy relations shown here
are the same.
At the extreme points of the oscillation, |x | = A , the KE is zero and the total mechanical
energy is equal to the PE. In the special case of SHM the restoring force is proportional to the
displacement so the force-displacement curve is a straight line (see §5-6, figure 5.6). From the area
under the curve, a triangle, the PE when x = A is:
FE7: Oscillations
78
maximum PE = 12 A (kA) = 12 kA2 = 12 m ω N2A2 .
The energy of the oscillation depends on the square of the amplitude A. This result is true for
all SHMs.
7-3
DAMPED OSCILLATIONS
If the total mechanical energy of an oscillating system were conserved, the system would oscillate
indefinitely with the same amplitude. In any real situation, however, there are always nonconservative forces such as friction or drag forces present. These dissipative forces decrease the
system's total mechanical energy and can be either internal or external to the system. The amplitude
will gradually decrease and the oscillations will die out. An oscillation which dies away is an
example of a transient motion. Examples include pendulums with damping forces.
Displacement
Time
Figure 7.5
A damped oscillation
The amplitude decreases with time and the system loses mechanical energy.
The frequency, fD, of a damped system is always less than fN, the natural frequency that the
system would have if the damping forces could be removed, since the damping forces always act to
retard the motion.
7-4
FORCED OSCILLATIONS
When non-conservative forces are present, an oscillation can die out unless energy is continually
supplied to the system by a driving force. An oscillation which is kept going by a periodic driving
force is called a forced oscillation. An example is a car engine.
If the driving force is sinusoidal (i.e. if a graph of the driving force against time looks like a
sine curve, perhaps shifted along the time axis), there will be a steady state oscillation set up at the
frequency, fF, of the driving force. In general, fF will be quite different from fD.
For a given amplitude of driving force, the amplitude of the steady state oscillation depends on
the driving frequency fF. An example is a vibrating metal rule driven by an electromagnet.
Each time the frequency of the driving force is changed, the new oscillation takes a while to
become stable, while the transient oscillations die away. As the driving frequency is increased from
zero, the amplitude of the steady state oscillation gradually increases until it reaches a peak and then
dies away again (figure 7.6). This phenomenon is called resonance. The frequency (fF) of the
driving force at which the amplitude of the oscillation is a maximum is called the resonance
frequency, fR, and the region of the graph near fR is called the resonance peak.
FE7: Oscillations
79
Resonance peak
Amplitude of
the steady state
oscillation
f
f
F
R
Figure 7.6 Response curve for a resonance
For lightly damped systems, fN , f D and fR are all approximately equal. At resonance the
driving force adds to the restoring force in the damped system, producing large accelerations and
oscillations.
Examples of forced oscillations
•
Shivering to heat the body in response to cold.
•
Expansion of the chest in breathing.
•
Beating of the heart and dilation and contraction of the arteries.
In each case the driving force is provided by the muscles. When these systems relax from
high to low potential energy states, their mechanical energy is decreased by the work done by nonconservative forces. The systems are highly damped and require the driving force to cause the
motion.
POST-LECTURE
7-5
QUESTIONS
Q7.2
List the forces that are acting on the following oscillating objects and in each case state whether the forces
are restoring, damping or driving forces:
i)
a tree swaying in the breeze,
ii)
the earth vibrating after an earthquake,
iii) a child on a swing,
iv) a vibrating eardrum,
v)
a marker buoy bobbing in the water,
vi) the atoms in a solid when a sound wave passes through.
Simple harmonic motion
In general one can obtain an order of magnitude estimate for the square of the natural angular
frequency (ω) of an oscillation by dividing an order of magnitude estimate of the restoring force per
FE7: Oscillations
80
displacement (k) by the mass (m) of the object in motion. This relation is exact for SHM (see
equation 7.4) but it also gives results which are roughly OK for other systems.
Q7.3
When a spring is stretched by a pair of 2.4 N forces its length increases by 62 mm. Predict the period of
oscillation when an object of mass 0.32 kg is hung from the spring and set vibrating. What will happen
to the period if the object is replaced by one with 10 times the mass?
Q7.4
A wooden rectangular pontoon of density
ρ = 0.50 × 1 03 kg.m-3 floats in water
(density ρW = 1.00 × 1 03 kg.m-3) so that two
faces, each of area A = 2 0 m2 , are horizontal.
The vertical sides each have a length L = l.0 m.
Figure 7.7
i)
How far is the bottom face below the water at equilibrium? (You may neglect the weight of the displaced
air in this problem.)
ii)
What is the total force acting on the pontoon when it is a small distance x
position with the same two faces horizontal?
below the equilibrium
iii) What would this force be if the pontoon were a distance x above the equilibrium position?
iv) Hence show that if the pontoon were released from a position other than its equilibrium position it would
execute SHM in the vertical direction if no non-conservative forces were present. (What non-conservative
forces would you expect to be present and what effect would they have?)
v)
Q7.5
What is the natural frequency of the above SHM ?
Use energy considerations to find the maximum speed of an object undergoing SHM in terms of the
amplitude and natural (angular) frequency of the motion.
Q7.6
A somewhat bizarre proposal for a rapid
transit system between Boston and Washington
Boston
Washington
(a distance of 630 km) made use of the Earth's
gravitational attraction. (Scientific American,
August 1965, pp 30 - 40.) Stripped of its
technical complications, the proposal had a
Centre of
capsule travelling through an evacuated tunnel
the Earth
drilled in a straight line between those cities.
For the first half of its journey, the capsule
would gradually be getting closer to the centre of
Figure 7.8
the earth (i.e. sliding down an incline); for the
second half it would be travelling away from the
Earth's centre.
In the idealised case where non-conservative forces are neglected, the potential energies at the beginning
and the end of the journey would be equal. (In the actual proposal, the capsule was helped along by
compressed air to overcome the non-conservative forces.) In this idealised case, the acceleration of the
capsule would be
a = -C x
where x is the displacement from the midpoint of the journey and C
1.55 × 10-6 s-2. The capsule would therefore be undergoing part of a SHM.
7-6
i)
How long would the one way trip from Boston to Washington take ?
ii)
What would be the maximum speed of the capsule on the journey ?
is a constant equal to
FURTHER DISCUSSION OF RESONANCE
In the video lecture it was stated that the Tacoma Narrows Bridge collapse was caused by resonance.
The driving force in that case, however, was not sinusoidal nor did it oscillate at the frequency of the
steady state oscillations of the bridge. In fact the driving force was provided by a wind blowing in
one direction for a time long compared to the period of the bridge's oscillation. How then does
resonance occur ?
FE7: Oscillations
81
Any driving force whatever may be considered mathematically to be a combination of
sinusoidal driving forces with appropriate frequencies, amplitudes and initial phases. The procedure
for splitting up a force in this way is called Fourier analysis (after the French mathematical
physicist who developed it to study the flow of heat through metals). Each of these sinusoidal
forces can be thought of as acting on the damped oscillating system independently of the other
sinusoidal components setting up a sinusoidal steady state oscillation at its own frequency. These
oscillations add up to give the actual oscillation caused by the full original driving force.
Schematically, this procedure can be summarised as shown in figure 7.9.
Full
driving
force
Full
oscillation
≡
+
+
+
+
...
≡
...
is equivalent to
+
+
Sum of sinusoidal
driving forces
Figure 7.9
Sum of sinusoidal
steady state oscillations
Resonant response to a complex driving force
The ratio of the amplitude of a steady state oscillation to that of the sinusoidal driving force
that caused it is largest when the driving force has a frequency equal to the resonance frequency.
Quite often therefore, the combined oscillation is dominated by the steady state oscillations with
frequency approximately equal to fR, particularly if the resonance curve is very sharply peaked.
These ideas are the basis of radio and TV receivers, for example. The driving force is the
complicated electrical signal (the sum of all the signals sent out by the various stations plus
interference from fluorescent lights, spark plugs etc.) from the aerial or antenna. The tuning circuit
of a radio or TV set is an oscillating system whose resonance frequency can be varied to select the
steady state oscillations at the frequency of the transmitting station. The resonance peak for these
tuning circuits has to be sufficiently narrow that only one station is transmitting at a frequency
within the peak at any given setting.
In the bridge collapse, the driving force provided by the wind had a large sinusoidal
component at one of the natural frequencies of oscillation of the bridge. This component caused the
bridge to resonate and ultimately break apart.
Q7.7
Can you think of any other situations in which resonance plays an important role ?
FE7: Oscillations
7-7 APPENDIX:
LISSAJOUS FIGURES
When two SHMs at right angles are combined, the path traced out will be a Lissajous figure if the
ratio of the two frequencies is equal to a ratio of two integers. The shape is also affected by the
phase difference between the oscillations.
1
1/2
1/3
2/3
3/4
3/5
4/5
5/6
Figure 7.10
Some Lissajous figures
The number at the left of each row is the ratio of the frequency of the vertical oscillation to that of the
horizontal oscillation for the figures in that row. The figures in each row correspond to different initial
phases of these two oscillations.
82
83
SCALE
FE8
OBJECTIVES
Aims
From this chapter you should learn to appreciate the great power of physical reasoning using very
rough approximations to more exact principles. You should become reasonably adept at applying
this kind of reasoning and the physical principles that you already know to problems of scaling comparing the behaviour of things, including animals, which have roughly the same shape and
composition.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the concept of scale factor.
2.
Explain how the laws of physics can be applied to a wide range of similar problems by
concentrating on the similarities and ignoring particular idiosyncrasies.
3.
Apply the concept of scale factor to analyse the way the sizes of different animals determine
their ability to perform simple mechanical operations such as running and jumping.
4.
Describe and explain the idea that bone strengths and muscular forces per cross-sectional area
are much the same for all animals.
5.
Recall and use the fact that the rate of energy expenditure by an animal is equal to the rate of
energy supply, which is controlled by the amount of oxygen available to the cells.
6.
State and apply the scaling laws listed in the table of §8-7.
7.
Describe some of the pitfalls to be avoided in scaling problems.
PRE-LECTURE
8-1
INTRODUCTION
The earlier parts of this unit have mainly been concerned with the application of physical concepts to
exact calculations, using carefully measured data. This technique is used, for example, by engineers
in designing bridges and power stations, by physicians in radiation dosimetry and by optometrists in
grinding lenses.
There is an equally important, although less well-known, way of using the concepts of physics
in non-exact calculations just to see whether a particular idea is valid. The arithmetic used may be
rough and ready - of the 3 × 3 = 10 variety. Frequent use may be made of statements involving the
words ‘about equal’. This is the style of approach used, for example, in forecasting trends in world
population and consumption of raw materials.
This second type of application of physics is the theme of this chapter. The concepts of forces
in equilibrium, work done by forces and energy balances will be used to examine some of the
features of the animal world - the strengths of skeletons, running, jumping and diving. The interest
is not in the subtler details of the mechanisms of these things, e.g. how a grasshopper bends its legs
prior to jumping or how a whale stocks up on oxygen before diving, but in broader considerations,
e.g. why grasshoppers, frogs and kangaroos all raise their centres of gravity by roughly the same
amount when jumping, or what advantages large mammals have over small mammals in diving.
FE8: Scale
In the spirit of the ‘equality’ 3 × 3 = 10, the physics used will be stripped to its bare
essentials. For the purposes of this chapter a force whose average magnitude is F moving through a
displacement d does work Fd - we will neglect niceties like including the cosine of the angle
between the force and the displacement. All such extraneous considerations will be lumped together
in proportionality constants which will have no bearing on the conclusions we draw from the
physics.
The primary aim of this chapter is to show how these perfectly respectable procedures can be
used to extract information. The results obtained simply illustrate the power of the method; you are
not expected to memorise these results for examination purposes. You will be expected to be able to
reason along the same lines.
Q8.1
There are two solid cylinders made of the same material. All the linear dimensions of the larger one are
exactly twice those of the smaller.
Compare their heights, diameters, base areas, curved surface areas, volumes and masses.
LECTURE
8-2
SCALE FACTOR
The question, ‘how much bigger is one tree than another?’ has no unique answer unless we know
on what basis the comparison is to be made - on their heights, on the cross-sectional area of their
trunks or on the volume of wood in their trunks, for example.
If the trees have roughly the same shape, a comparison can be made in terms of a measure
based on length - the scale factor, L. Then if one tree were L times taller than the other, the radius
of its trunk would be L times bigger, the cross-sectional area of its trunk L2 times bigger and the
volume of its trunk L3 times bigger than the other. The symbol L has been chosen deliberately as a
reminder that our basic comparison is being made on corresponding lengths. However L is a ratio
of lengths; it has no unit; it is a scale factor, not a scale length.
A scale factor can be assigned to any class of objects which have essentially the same shape.
In the following we will be comparing the properties of different species of animals which have
roughly similar basic shapes, but which differ in size, e.g. dogs, sheep and horses.
8-3
BONE LOADS AND MUSCULAR FORCES
We make the assumptions that the strength of the materials that bones are made of and the way the
bones are constructed are much the same for all animals. By strength we mean the maximum
magnitude of the force per cross-sectional area that the bone is able to withstand under a
compressional or an extensive load (the two values need not be the same). This means that the
maximum force that can be applied along the length of a bone without breaking it depends on the
cross-sectional area of the bone, i.e. on the square of the bone's diameter.
This assumption is supported indirectly by two sets of experimental results on the breaking of
tibia bones of different sizes. (Notice that in this experiment the loads are applied at right angles to
the bones - they are not along the bones themselves.)
84
FE8: Scale
85
•
In the experiment shown on TV, 37 force units were required to break a dog's tibia of diameter
1.66 length units while 50 force units were required to break a sheep's tibia of diameter 1.84 length
units.
•
The following results were obtained for the tibia bones of different breeds of dog (figure 8.1).
2.0
Diameter/
arbitrary
unit
1.5
•
•
20
30
•• • •
•
•
• • • •• •
• • • ••••
•
•• ••
•
1.0
•
• •
•
•
•
40
Force/arbitrary unit
Figure 8.1 Force required to break a dog's bone
The same type of assumptions about the maximum forces supplied by muscles or tendons
lead to the result that these forces also scale as the square of the diameter.
Applications
Animals standing still
The leg bones of a stationary animal must support a compressional load proportional to the animal's
weight. The weight of the animal is proportional to its volume and so, between different sized
animals, varies as L3 . We saw above that the maximum compressional force which the bones can
support varies as the square of the diameter. The diameter of the leg bones of similar animals must
therefore increase faster than the other linear dimensions of these animals. Thus, for example, the
leg bones of elephants are comparatively much thicker than those of mice.
Moving animals
The legs of an animal are also subject to bending forces, particularly when they are moving.
Consider for example the forces acting on the humerus bone of a mammal.
T
F
F
T
Figure 8.2 Forces on a bone
The adjacent bones exert vertical forces F proportional to the weight of the mammal (to L3).
To keep the bone from twisting (rotating) there must be a pair of equal and opposite forces T
FE8: Scale
86
exerted by the muscles and the tendons along the humerus bone so that there is no net torque: these
forces provide a bending (or shearing) load on the bone.
d
l
Figure 8.3
so
Dimensions of a bone
If the length of the humerus bone is l and its diameter is d, since there is no net torque,
Fl - T d = 0
Fl
T = d .
Since l and d both vary as L , then T , like F , varies as L3 .
The bending forces T act over areas A near the surface of the bone as shown in the crosssectional view below. These areas, and hence the capacity of the bone to withstand the bending, vary
as the square of the diameter.
A
bone
d
marrow
A
Figure 8.4
Location of bending forces in a leg bone
Thus the bending forces in the humerus increase more rapidly with L than the capacity of the
bone to withstand them - essentially the same result as in the case of standing still. These bending
forces are the reason why giraffes keep their legs straight while running.
8-4
SUPPLY OF CHEMICAL ENERGY IN THE BODY
The maximum continuous rate at which an animal can do work while engaging in various activities is
limited by the rate at which the cells in the body can supply energy to the muscles. The latter rate is
controlled by the amount of oxygen available to the cells; it is shown in the POST-LECTURE to vary
as L2 .
Thus, for animals of similar shape,
the maximum rate of energy supply = k1L2 ,
.... (8.1)
where k1 is a constant independent of the size of the animal and hence of L.
Applications
Running on the flat
Does the size of an animal determine the speed at which it can run on the flat? While the animal is
running at a fixed speed, here its maximum speed v, its kinetic energy is constant. Since it is
FE8: Scale
87
running on flat ground its gravitational potential energy is also constant. Hence the energy supplied
(by the cells) and the (negative) work being done by the drag force exerted by the air must total zero.
The drag force is proportional to the cross-sectional area of the animal and to the square of the
speed. In symbols
drag force = k2L2 v2 ,
where k2 is independent of L and v .
In a time interval t the animal travels a distance vt so the work done by the drag force during
this time is -k2L2 v3 t.
Equating the magnitude of the rate of doing this work with the maximum rate of energy supply
(from equation 1 above),
k2L2 v3 = klL2 ,
v3
or
=
k1
k2
= constant.
This means that, on the basis of this scaling argument, the maximum speed of an animal
running on the flat is expected to be independent of its size. This expectation agrees with the
observation that, whilst there are slight individual differences between the maximum speeds of
animals, there is no systematic trend with size. For example, dogs and horses can run at about
16 m.s-1 while humans, intermediate in size between these, run at about 10 m.s-1.
Running up a steep incline
The dominant force doing work on the animal is the gravitational force. This force (the weight) is
proportional to L3 . As in the previous application, if the animal is travelling at constant speed the
energy supplied (by the cells) and the work being done by the gravitational force must total zero.
Suppose that the animal is travelling at its maximum speed v up the incline. In a time interval t
the animal travels a distance vt so the work done by the gravitational force during this time is -k3 L3 vt
. (k3 is also independent of L and v .)
Equating the magnitude of the rate of doing this work to the maximum rate of energy supply,
k3 L3v = k1 L2
or
v
=
kl -1
k3 L .
In other words, the maximum speed of an animal travelling up an incline would be expected to
be inversely proportional to its size. For example a greyhound would be expected to travel faster
than a horse up a hill.
8-5
JUMPING
An animal standing on the ground exerts a force on the ground equal in magnitude to the weight of
the animal. In order to jump, the animal first crouches, lowering its centre of gravity by a small
displacement d. It then pushes against the ground with an additional force of average magnitude F.
This means that there is an unbalanced force of this average magnitude upwards on the animal as
long as it remains in contact with the ground. As the animal leaves the ground it accelerates upwards
and its kinetic energy increases as its centre of gravity returns to its original position. This increase
in KE actually comes from work done by the leg muscles but its value can be calculated using the
product of the net external force (F) and the displacement (d) of the centre of gravity. So the work
done on the animal is equal to Fd.
After the animal leaves the ground its centre of gravity rises to a height h above its position
when it was crouching.
FE8: Scale
88
h
d
Figure 8.5
Location of the centre of gravity during a jump
The associated increase in gravitational potential energy can be equated to the kinetic energy at
take-off or the work done during the launch.
The increase in gravitational potential energy is k5 L3 h where k5 is independent of L and h.
The average force F provided by the muscles is proportional to L2 (see earlier) and the
crouching distance d is proportional to L, so the work done is equal to k4L3 where k4 is independent
of L .
Equating these two quantities:
k5 L3h = k 4L3 ;
hence
h
=
k4
k5 .
This means that the maximum height through which the centre of gravity can be raised is
expected to be independent of the animal's size. Observation shows that this is so; for most animals
the height involved is a little over 1 metre.
8-6
DIVING
When a mammal dives below the surface of the water, it consumes oxygen at a rate proportional to
L2 (see the section in the post-lecture on the rate of supply of energy); in a time interval t the
mammal consumes a quantity k6 L2 t of oxygen (k6 is independent of L and t).
The total oxygen supply depends on the volume of the lungs and hence can be written as k7 L3 .
(k7 is independent of L.) Hence the maximum duration of the dive is given by
k7
t = k L,
6
which, since k6 and k7 are independent of L, indicates that larger mammals can stay under water for
longer periods. Again this is confirmed by observation.
POST-LECTURE
8-7
APPLICATION OF THE CONCEPT OF SCALE FACTOR
Applications involving scale factors fall into two classes.
•
The first, rather straight-forward, class is that in which measured or given data lead directly or
indirectly to a value of the scale factor L and this value can be used to estimate other data pertinent to
the system. In dealing with these problems it is useful to be able to recollect the way certain
fundamental quantities are related to the scale factor. Some of these relations are summarised here
in tabular form.
FE8: Scale
L
89
L2
surface area
cross-sectional area
rate of energy supply
etc.
length
girth
height
etc.
L3
volume
mass
weight
etc.
An example of this class of problem is as follows. Consider two oranges, both roughly
spherical, one of mass 0.20 kg, the other of mass 0.40 kg and ask the question as to how the surface
areas of the two oranges compare. To answer this one uses the mass ratio (2:1) to find the value of
L3 and proceeds to compare surface areas through the derived value of L2 .
•
The second class is where it is required to derive the way in which the variation of some
quantity among things of similar shape is related to the scale factor. Such problems usually are
more involved and require the use of some linking factor known from everyday life or from the
formal study of physics. Many of the illustrations in the lecture material were of this type: running,
jumping and diving animals, to name a few.
Q8.2
Suppose that in order to make Australians champions in all sporting events we breed fair dinkum sunbronzed Aussies who are scaled up × 2 in all linear dimensions. They are made of the same stuff as
conventional × 1 people. How would a × 2 person's performance compare with that of a × 1 person in
the following events:
a) a form of high jumping where, to even things up, the judges measure the increases in heights of the
jumpers' centres of gravity,
b) weight-lifting,
c) sprinting (assume that the forward acceleration from rest at the beginning of the sprint is the dominant
influence on the sprinters' performances),
d) putting a shot made of the usual material but whose linear dimensions are proportional to the height of the
shot putter? (For simplicity assume that the shot is launched horizontally.)
8-8
BREAKING OF DOGS' BONES
The experimental results shown in the lecture for the tibia bones of different breeds of dog are
reproduced in graphical form below.
•
Force/
arbitrary
unit
40
•
30
•
• •
•
• ••
•
•••• ••
••
•
••
• •• •
••• • •
•• •
20
10
0 •
0
Figure 8.6
5
10
15
20
Diameter/arbitrary unit
Linear plot of the force required to break a dog's bone
This graph differs from figure 8.1 in that the origin is shown on the graph. This point is a
data point on logical grounds rather than experimental evidence. You can now see that the
relationship between breaking force and bone diameter is not linear. It may follow some sort of
power law. The axes of the graph have been swapped to make the following question simpler.
FE8: Scale
Q8.3
90
Suppose the relationship between diameter d and force F is of the form
F= kdn
where k and n are constants.
Re-plot the data on a log-log graph to see if the data support the view, held in the lecture, that n = 2.
5
4
3
2
1
9
8
7
6
5
5
Figure 8.7
6 7 8 9 1
2
3
4
5
Logarithmic plot of the force needed to break a dog's bone
You should replot the data from figure 8.6.
8-8
RATE OF ENERGY SUPPLY AND PULSE RATE
Rate of energy supply in an animal
The rate of supply of energy to muscles depends on the rate of oxygen supply which, in turn,
depends on the volume rate of flow of blood passing through the muscle tissues. This volume rate
of flow of blood depends on the cross-sectional area of the aorta (proportional to L2 ) and the speed
of the blood flow.
The first question then is whether the speed of the blood flow depends on size. This question
can be answered in a round-about fashion. Since the speed of the blood flow is itself proportional to
the blood pressure the question becomes: does the blood pressure depend on size?
A
1
A
Figure 8.8
2
Cross-section through an artery
The muscular action in the ventricle of the heart keeps the blood moving with more or less
constant volume rate of flow so the force exerted by the muscles equals that arising from the
pressure. If P is the blood pressure, S the muscular force per cross-sectional area of the muscles, Al
is the cross-sectional area of the ventricle and A2 is the cross-sectional area of the muscle,
P A1 = S A2 ,
that is
P
=
A2
S A .
l
FE8: Scale
91
The two areas depend on L2 while S is independent of L. This means that the pressure, and
hence the speed of blood through the aorta, does not depend on size.
Therefore both the rate of flow of blood delivered by the heart and the rate at which energy is
supplied are proportional to the cross-sectional area of the aorta and hence depend on L2 .
Pulse rate
Q8.4
We have seen that the rate of blood supply from the heart varies as L2 . The volume of the heart will
naturally vary as L3 . What conclusion can you reach regarding the pulse rate of animals of various sizes?
8-9
A CAUTIONARY TALE ABOUT DRUG DOSAGES
Scaling problems can often present traps for the unwary, even for experimenters who should know
better. A few years ago some scientists wished to test the reaction of elephants to LSD dosage.
They calculated the dose from that already found to put a cat into a rage, (0.1 mg of LSD per kg of
body weight) and scaled up by weight to get a dose of 297 mg. On receiving this dose the elephant
immediately started trumpeting and running around, then it stopped, swayed and, after five minutes,
collapsed, went into convulsions, defecated and died. The scientists concluded that elephants are
peculiarly sensitive to LSD.
Was the scaling done correctly? There are several approaches to this problem, leading to a
large range of suggested doses.
(a) Equal concentrations of LSD in body fluids for elephant and cat, i.e. scaling according to
weight. Dose: 297 mg.
(b) Scaling based on metabolic rate, which will control detoxification of the drug and its excretion.
Dose: 80 mg.
(c) Scaling based on an animal which is not as notoriously tolerant to LSD as a cat, for example
we could use a human. A dose of 0.2 mg LSD produces severe psychotic symptoms in
humans. Using a scaling based on weight, the elephant dose is 8 mg.
(d) Scaling on metabolic rate, using the human dose as a base. Dose: 3 mg.
(e) Scaling on brain size, since LSD could be more concentrated there, again using human dose as
base (human brain 1.4 kg, elephant brain 3 kg ). Dose: 0.4 mg.
8-10
EARLY ATTEMPTS AT SCALING
The idea of scaling in physics is not new. You may be interested in reading the discussion in
Galileo's work Concerning the Two New Sciences, pp 184-188, vol 28, Britannica Great Books
which is available in the Fisher Library Reading Room. Jonathan Swift encountered the problem in
Gulliver's Travels where the Lilliputian emperor decided that Gulliver needed (12)3 , that is 1728,
Lilliputian portions of food. Actually the emperor overestimated the situation by scaling according
to weight rather than to surface area.
8-11
SCALING APPLIED TO MOTOR VEHICLES
A feature of scaling problems involving animals is that the power available varies as L2 . The
builders of machines are not bound by this constraint and can adjust the scaling factors to best suit
their requirements. For example, as the size of vehicles is increased from cars to semi-trailers the
prime consideration is to increase the carrying capacity, even if the motive power does not increase in
the same proportion as in the animal world.
Q8.5
A family car is about 3 m long, has a mass of about 1000 kg and develops a power of about 75 kW. A
semitrailer is about 9 m long, has a laden mass of about 30000 kg and develops a power of about 225 kW.
How much faster can the car climb a hill than the semitrailer ?